From 11 all operations with fractions. Complex expressions with fractions

1. The rule for adding fractions with the same denominators:

Example 1:

Example 2:

Rule for adding fractions with different denominators:

Example 1:

Example 2:

Here the denominators were not multiplied, but the least common factor a2 was taken.
(The denominator has the highest power of 2.)
An additional factor for the first fraction is 1, for the second it is a.

2. Rule for subtracting fractions with like denominators:

Rule for subtracting fractions with different denominators:

3. Rule for multiplying ordinary fractions:

4. Rule for dividing fractions:

Example:

Ordinary (simple) fraction. Numerator and denominator of a fraction.
Proper and improper fractions. Mixed number.
Incomplete quotient. Integer and fractional parts. Reverse fractions. Part of a unit or several parts of it are called ordinary or simple fraction. Quantity equal parts, into which the unit is divided is called the denominator, and the number of parts taken is called the numerator. The fraction is written as:

Here 3 is the numerator, 7 is the denominator.

If the numerator is less than the denominator, then the fraction is less than 1 and is called proper fraction. If the numerator is equal to the denominator, then the fraction is equal to 1. If the numerator is greater than the denominator, then the fraction is greater than 1. In both of the latter cases, the fraction is called improper. If the numerator is divided by the denominator, then this fraction is equal to the quotient of the division: 63 / 7 = 9. If the division is performed with a remainder, then this improper fraction can be represented mixed number:

Here 9 – incomplete quotient (whole part mixed number), 2 – remainder (numerator of the fractional part), 7 – denominator.
It is often necessary to solve the inverse problem - reverse a mixed number into a fraction. To do this, multiply the integer part of the mixed number by the denominator and add the numerator of the fractional part. This will be the numerator of the common fraction, but the denominator remains the same.

Reciprocal fractions are two fractions whose product is equal to 1. For example, 3 / 7 and 7 / 3; 15/1 and 1/15, etc.

Fraction expansion. Reducing a fraction. Comparing fractions.
Reduction to a common denominator. Addition and subtraction fractions.
Multiplying fractions. Division of fractions
Fraction expansion.The value of a fraction does not change if its numerator and denominator are multiplied by the same number other than zero by expanding the fraction. For example,

Reducing a fraction. The value of a fraction does not change if you divide its numerator and denominator by the same number other than zero.. This transformation is calledreducing a fraction. For example,

Comparing fractions.Of two fractions with the same numerators, the one whose denominator is smaller is greater:

Of two fractions with the same denominator, the one whose numerator is greater is greater:

To compare fractions that have different numerators and denominators, you need to expand them to bring them to a common denominator.
EXAMPLE Compare two fractions:

The transformation used here is called bringing fractions to a common denominator.
Adding and subtracting fractions.If the denominators of the fractions are the same, then in order to add the fractions, you need to add their numerators, and in order to subtract the fractions, you need to subtract their numerators (in the same order). The resulting sum or difference will be the numerator of the result; the denominator will remain the same. If the denominators of the fractions are different, you must first reduce the fractions to a common denominator. When adding mixed numbers, their whole and fractional parts are added separately. When subtracting mixed numbers, we recommend first converting them to improper fractions, then subtracting one from the other, and then converting the result again, if necessary, to mixed number form.
EXAMPLE

Multiplying fractions.To multiply a number by a fraction means to multiply it by the numerator and divide the product by the denominator. Therefore we have general rule multiplying fractions:to multiply fractions, you need to multiply their numerators and denominators separately and divide the first product by the second.
EXAMPLE
Dividing fractions. In order to divide a certain number by a fraction, it is necessary to multiply this number by the reciprocal fraction. This rule follows from the definition of division (see the section “Arithmetic operations”).
EXAMPLE

Decimal. The whole part. Decimal point.
Decimal places. Properties of decimal fractions.
Periodic decimal fraction. Period
Decimalis the result of dividing one by ten, one hundred, thousand, etc. parts. These fractions are very convenient for calculations, since they are based on the same positional system on which counting and writing integers are based. Thanks to this, the notation and rules for working with decimal fractions are essentially the same as for whole numbers. When writing decimal fractions, there is no need to mark the denominator; this is determined by the place occupied by the corresponding digit. First it is written whole part numbers, then put on the rightdecimal point. The first digit after the decimal point means the number of tenths, the second – the number of hundredths, the third – the number of thousandths, etc. The numbers located after the decimal point are calleddecimals.
EXAMPLE
One of the advantages of decimal fractions is that they are easily reduced to ordinary fractions: the number after the decimal point (in our case 5047) is the numerator; the denominator is equal n -th power of 10, where n - number of decimal places (in our case n = 4):
If the decimal fraction does not contain an integer part, then a zero is placed before the decimal point:

Properties of decimal fractions.

1. The decimal does not change if you add zeros to the right:

2. The decimal fraction does not change if you remove the zeros located
at the end of the decimal:

0.00123000 = 0.00123 .

Attention! You cannot remove zeros that are not located at the end decimal!br />

These properties allow you to quickly multiply and divide decimals by 10, 100, 1000, etc.

Periodic decimal contains an infinitely repeating group of numbers called a period. The period is written in brackets. For example, 0.12345123451234512345… = 0.(12345).

EXAMPLE If we divide 47 by 11, we get 4.27272727... = 4.(27).

Multiplying decimals.
Dividing decimals.

Adding and subtracting decimals. These operations are performed in the same way as adding and subtracting integers. You just need to write down the corresponding decimal places one below the other.
EXAMPLE

Multiplying decimals. At the first stage, we multiply decimal fractions as whole numbers, without taking into account the decimal point. Then the following rule applies: the number of decimal places in the product is equal to the sum of the decimal places in all factors.
Note: before placing the decimal point inthe product cannot be discarded with trailing zeros!
EXAMPLE

The sum of the numbers of decimal places in the factors is equal to: 3 + 4 = 7. The sum of the numbers in the product is 6. Therefore, you need to add one zero to the left: 0197056 and put a decimal point in front of it: 0.0197056.
Decimal division
Dividing a decimal by a whole number
If dividend is less than divisor, write a zero in the integer part of the quotient and put a decimal point after it. Then, without taking into account the decimal point of the dividend, we add the next digit of the fractional part to its integer part and again compare the resulting integer part of the dividend with the divisor. If the new number is again less than the divisor, we put another zero after the decimal point in the quotient and add the next digit of its fractional part to the whole part of the dividend. We repeat this process until the resulting dividend becomes greater than the divisor. After this, division is performed as for integers. If the dividend is greater than or equal to the divisor, first we divide its whole part, write the result of division in the quotient and put a decimal point. After this, division continues as in the case of integers.
EXAMPLE Divide 1.328 by 64.
Solution:
Dividing one decimal fraction by another.
First, we transfer the decimal points in the dividend and divisor to the number of decimal places in the divisor, that is, we make the divisor an integer. Now we perform division as in the previous case.
EXAMPLE Divide 0.04569 by 0.0006.
Solution: Move the decimal points 4 positions to the right and divide 456.9 by 6:

In order to reverse decimal in an ordinary one, you need to take the number after the decimal point as the numerator, and take the nth power of ten as the denominator (here n is the number of decimal places). The non-zero integer part is stored in an ordinary fraction; the zero integer part is omitted. For example:
In order to convert a fraction to a decimal, you must divide the numerator by the denominator in accordance with the division rules.
EXAMPLE Convert 5/8 to a decimal.
Solution: Dividing 5 by 8 gives 0.625. (Check, please!).
In most cases, this process can continue indefinitely. Then it is impossible to accurately convert a fraction to a decimal. But in practice this is never required. The division is aborted if the decimal places of interest have already been obtained.
EXAMPLE Convert 1/3 to a decimal.
Solution: Dividing 1 by 3 will be infinite: 1:3 = 0.3333… .
Check it out please!

496. Find X, If:

497. 1) If you add 10 1/2 to 3/10 of an unknown number, you get 13 1/2. Find the unknown number.

2) If you subtract 10 1/2 from 7/10 of an unknown number, you get 15 2/5. Find the unknown number.

498 *. If you subtract 10 from 3/4 of an unknown number and multiply the resulting difference by 5, you get 100. Find the number.

499 *. If you increase an unknown number by 2/3 of it, you get 60. What number is this?

500 *. If you add the same amount to the unknown number, and also 20 1/3, you get 105 2/5. Find the unknown number.

501. 1) The potato yield with square-cluster planting averages 150 centners per hectare, and with conventional planting it is 3/5 of this amount. How much more potatoes can be harvested from an area of ​​15 hectares if potatoes are planted using the square-cluster method?

2) An experienced worker produced 18 parts in 1 hour, and an inexperienced worker produced 2/3 of this amount. How many more parts can an experienced worker produce in a 7-hour day?

502. 1) The pioneers collected 56 kg of different seeds over three days. On the first day, 3/14 of the total amount was collected, on the second, one and a half times more, and on the third day, the rest of the grain. How many kilograms of seeds did the pioneers collect on the third day?

2) When grinding the wheat, the result was: flour 4/5 of the total amount of wheat, semolina - 40 times less than flour, and the rest was bran. How much flour, semolina and bran separately were produced when grinding 3 tons of wheat?

503. 1) Three garages can accommodate 460 cars. The number of cars that fit in the first garage is 3/4 of the number of cars that fit in the second, and the third garage has 1 1/2 times as many cars as the first. How many cars fit in each garage?

2) A factory with three workshops employs 6,000 workers. In the second workshop there are 1 1/2 times fewer workers than in the first, and the number of workers in the third workshop is 5/6 of the number of workers in the second workshop. How many workers are there in each workshop?

504. 1) First 2/5, then 1/3 of the total kerosene was poured from a tank with kerosene, and after that 8 tons of kerosene remained in the tank. How much kerosene was in the tank initially?

2) The cyclists raced for three days. On the first day they covered 4/15 of the entire journey, on the second - 2/5, and on the third day the remaining 100 km. How far did the cyclists travel in three days?

505. 1) The icebreaker fought its way through the ice field for three days. On the first day he walked 1/2 of the entire distance, on the second day 3/5 of the remaining distance and on the third day the remaining 24 km. Find the length of the path covered by the icebreaker in three days.

2) Three groups of schoolchildren planted trees to green the village. The first detachment planted 7/20 of all trees, the second 5/8 of the remaining trees, and the third the remaining 195 trees. How many trees did the three teams plant in total?

506. 1) A combine harvester harvested wheat from one plot in three days. On the first day, he harvested from 5/18 of the entire area of ​​the plot, on the second day from 7/13 of the remaining area, and on the third day from the remaining area of ​​30 1/2 hectares. On average, 20 centners of wheat were harvested from each hectare. How much wheat was harvested in the entire area?

2) On the first day, the rally participants covered 3/11 of the entire route, on the second day 7/20 of the remaining route, on the third day 5/13 of the new remainder, and on the fourth day the remaining 320 km. How long is the route of the rally?

507. 1) On the first day the car covered 3/8 of the entire distance, on the second day 15/17 of what it covered on the first, and on the third day the remaining 200 km. How much gasoline was consumed if a car consumes 1 3/5 kg of gasoline for 10 km?

2) The city consists of four districts. And 4/13 of all residents of the city live in the first district, 5/6 of the residents of the first district live in the second, 4/11 of the residents of the first live in the third; two districts combined, and 18 thousand people live in the fourth district. How much bread does the entire population of the city need for 3 days, if on average one person consumes 500 g per day?

508. 1) The tourist walked on the first day 10/31 of the entire journey, on the second 9/10 of what he walked on the first day, and on the third the rest of the way, and on the third day he walked 12 km more than on the second day. How many kilometers did the tourist walk on each of the three days?

2) The car covered the entire route from city A to city B in three days. On the first day the car covered 7/20 of the entire distance, on the second 8/13 of the remaining distance, and on the third day the car covered 72 km less than on the first day. What is the distance between cities A and B?

509. 1) The Executive Committee allocated land to the workers of three factories for garden plots. The first plant was allocated 9/25 of the total number of plots, the second plant 5/9 of the number of plots allocated for the first, and the third - the remaining plots. How many total plots were allocated to the workers of three factories, if the first factory was allocated 50 fewer plots than the third?

2) The plane delivered a shift of winter workers to the polar station from Moscow in three days. On the first day he flew 2/5 of the entire distance, on the second - 5/6 of the distance he covered on the first day, and on the third day he flew 500 km less than on the second day. How far did the plane fly in three days?

510. 1) The plant had three workshops. The number of workers in the first workshop is 2/5 of all workers in the plant; in the second workshop there are 1 1/2 times fewer workers than in the first, and in the third workshop there are 100 more workers than in the second. How many workers are there in the factory?

2) The collective farm includes residents of three neighboring villages. The number of families in the first village is 3/10 of all families on the collective farm; in the second village the number of families is 1 1/2 times greater than in the first, and in the third village the number of families is 420 less than in the second. How many families are there on the collective farm?

511. 1) The artel used up 1/3 of its stock of raw materials in the first week, and 1/3 of the rest in the second week. How much raw material is left in the artel if in the first week the consumption of raw materials was 3/5 tons more than in the second week?

2) Of the imported coal, 1/6 of it was spent for heating the house in the first month, and 3/8 of the remainder in the second month. How much coal is left to heat the house if 1 3/4 more was used in the second month than in the first month?

512. 3/5 of the total land of the collective farm is allocated for sowing grain, 13/36 of the remainder is occupied by vegetable gardens and meadows, the rest of the land is forest, and the sown area of ​​the collective farm is 217 hectares larger than the forest area, 1/3 of the land allocated for sowing grain is sown with rye, and the rest is wheat. How many hectares of land did the collective farm sow with wheat and how many with rye?

513. 1) The tram route is 14 3/8 km long. Along this route, the tram makes 18 stops, spending on average up to 1 1/6 minutes per stop. The average speed of the tram along the entire route is 12 1/2 km per hour. How long does it take for a tram to complete one trip?

2) Bus route 16 km. Along this route the bus makes 36 stops of 3/4 minutes each. on average each. The average bus speed is 30 km per hour. How long does a bus take for one route?

514*. 1) It’s 6 o’clock now. evenings. What part is the remaining part of the day from the past and what part of the day is left?

2) A steamer travels the distance between two cities with the current in 3 days. and back the same distance in 4 days. How many days will the rafts float downstream from one city to another?

515. 1) How many boards will be used to lay the floor in a room whose length is 6 2/3 m, width 5 1/4 m, if the length of each board is 6 2/3 m, and its width is 3/80 of the length?

2) A rectangular platform has a length of 45 1/2 m, and its width is 5/13 of its length. This area is bordered by a path 4/5 m wide. Find the area of ​​the path.

516. Find the average arithmetic numbers:

517. 1) The arithmetic mean of two numbers is 6 1/6. One of the numbers is 3 3/4. Find another number.

2) The arithmetic mean of two numbers is 14 1/4. One of these numbers is 15 5/6. Find another number.

518. 1) The freight train was on the road for three hours. In the first hour he covered 36 1/2 km, in the second 40 km and in the third 39 3/4 km. Find the average speed of the train.

2) The car traveled 81 1/2 km in the first two hours, and 95 km in the next 2 1/2 hours. How many kilometers did he walk on average per hour?

519. 1) The tractor driver completed the task of plowing the land in three days. On the first day he plowed 12 1/2 hectares, on the second day 15 3/4 hectares and on the third day 14 1/2 hectares. On average, how many hectares of land did a tractor driver plow per day?

2) A group of schoolchildren, making a three-day tourist trip, were on the road for 6 1/3 hours on the first day, 7 hours on the second. and on the third day - 4 2/3 hours. How many hours on average did schoolchildren travel every day?

520. 1) Three families live in the house. The first family has 3 light bulbs to illuminate the apartment, the second has 4 and the third has 5 light bulbs. How much should each family pay for electricity if all the lamps were the same, and the total electricity bill (for the whole house) was 7 1/5 rubles?

2) A polisher was polishing the floors in an apartment where three families lived. The first family had a living area of ​​36 1/2 square meters. m, the second is 24 1/2 sq. m, and the third - 43 sq. m. For all the work, 2 rubles were paid. 08 kop. How much did each family pay?

521. 1) In the garden plot, potatoes were collected from 50 bushes at 1 1/10 kg per bush, from 70 bushes at 4/5 kg per bush, from 80 bushes at 9/10 kg per bush. How many kilograms of potatoes are harvested on average from each bush?

2) The field crew on an area of ​​300 hectares received a harvest of 20 1/2 quintals of winter wheat per 1 hectare, from 80 hectares to 24 quintals per 1 ha, and from 20 hectares - 28 1/2 quintals per 1 ha. What is the average yield in a brigade with 1 hectare?

522. 1) The sum of two numbers is 7 1/2. One number is 4 4/5 greater than the other. Find these numbers.

2) If we add the numbers expressing the width of the Tatar and Kerch Straits together, we get 11 7/10 km. The Tatar Strait is 3 1/10 km wider than the Kerch Strait. What is the width of each strait?

523. 1) The sum of three numbers is 35 2 / 3. The first number is greater than the second by 5 1/3 and greater than the third by 3 5/6. Find these numbers.

2) Islands New Earth, Sakhalin and Severnaya Zemlya together occupy an area of ​​196 7/10 thousand square meters. km. The area of ​​Novaya Zemlya is 44 1/10 thousand square meters. km larger than the area of ​​Severnaya Zemlya and 5 1/5 thousand square meters. km larger than the area of ​​Sakhalin. What is the area of ​​each of the listed islands?

524. 1) The apartment consists of three rooms. The area of ​​the first room is 24 3/8 sq. m and is 13/36 of the entire area of ​​the apartment. The area of ​​the second room is 8 1/8 square meters. m more than the area of ​​the third. What is the area of ​​the second room?

2) A cyclist during a three-day competition on the first day was on the road for 3 1/4 hours, which was 13/43 of the total travel time. On the second day he rode 1 1/2 hours more than on the third day. How many hours did the cyclist travel on the second day of the competition?

525. Three pieces of iron weigh together 17 1/4 kg. If the weight of the first piece is reduced by 1 1/2 kg, the weight of the second by 2 1/4 kg, then all three pieces will have the same weight. How much did each piece of iron weigh?

526. 1) The sum of two numbers is 15 1/5. If the first number is reduced by 3 1/10, and the second is increased by 3 1/10, then these numbers will be equal. What is each number equal to?

2) There were 38 1/4 kg of cereal in two boxes. If you pour 4 3/4 kg of cereal from one box into another, then there will be equal amounts of cereal in both boxes. How much cereal is in each box?

527 . 1) The sum of two numbers is 17 17 / 30. If you subtract 5 1/2 from the first number and add it to the second, then the first will still be greater than the second by 2 17/30. Find both numbers.

2) There are 24 1/4 kg of apples in two boxes. If you transfer 3 1/2 kg from the first box to the second, then in the first there will still be 3/5 kg more apples than in the second. How many kilograms of apples are in each box?

528 *. 1) The sum of two numbers is 8 11/14, and their difference is 2 3/7. Find these numbers.

2) The boat moved along the river at a speed of 15 1/2 km per hour, and against the current at 8 1/4 km per hour. What is the speed of the river flow?

529. 1) There are 110 cars in two garages, and in one of them there are 1 1/5 times more than in the other. How many cars are in each garage?

2) The living area of ​​an apartment consisting of two rooms is 47 1/2 sq. m. m. The area of ​​one room is 8/11 of the area of ​​the other. Find the area of ​​each room.

530. 1) An alloy consisting of copper and silver weighs 330 g. The weight of copper in this alloy is 5/28 of the weight of silver. How much silver and how much copper is in the alloy?

2) The sum of two numbers is 6 3/4, and the quotient is 3 1/2. Find these numbers.

531. The sum of three numbers is 22 1/2. The second number is 3 1/2 times, and the third is 2 1/4 times the first. Find these numbers.

532. 1) The difference of two numbers is 7; quotient of division more for less 5 2/3. Find these numbers.

2) The difference between two numbers is 29 3/8, and their multiple ratio is 8 5/6. Find these numbers.

533. In a class, the number of absent students is 3/13 of the number of students present. How many students are in the class according to the list if there are 20 more people present than absent?

534. 1) The difference between two numbers is 3 1/5. One number is 5/7 of another. Find these numbers.

2) The father is 24 years older than his son. The number of the son's years is equal to 5/13 of the father's years. How old is the father and how old is the son?

535. The denominator of a fraction is 11 units greater than its numerator. What is the value of a fraction if its denominator is 3 3/4 times the numerator?

No. 536 - 537 orally.

536. 1) The first number is 1/2 of the second. How many times is the second number greater than the first?

2) The first number is 3/2 of the second. What part of the first number is the second number?

537. 1) 1/2 of the first number is equal to 1/3 of the second number. What part of the first number is the second number?

2) 2/3 of the first number is equal to 3/4 of the second number. What part of the first number is the second number? What part of the second number is the first?

538. 1) The sum of two numbers is 16. Find these numbers if 1/3 of the second number is equal to 1/5 of the first.

2) The sum of two numbers is 38. Find these numbers if 2/3 of the first number is equal to 3/5 of the second.

539 *. 1) Two boys collected 100 mushrooms together. 3/8 of the number of mushrooms collected by the first boy is numerically equal to 1/4 of the number of mushrooms collected by the second boy. How many mushrooms did each boy collect?

2) The institution employs 27 people. How many men work and how many women work if 2/5 of all men are equal to 3/5 of all women?

540 *. Three boys bought a volleyball. Determine the contribution of each boy, knowing that 1/2 of the contribution of the first boy is equal to 1/3 of the contribution of the second, or 1/4 of the contribution of the third, and that the contribution of the third boy is 64 kopecks more than the contribution of the first.

541 *. 1) One number is 6 more than the other. Find these numbers if 2/5 of one number are equal to 2/3 of the other.

2) The difference of two numbers is 35. Find these numbers if 1/3 of the first number is equal to 3/4 of the second number.

542. 1) The first team can complete some work in 36 days, and the second in 45 days. In how many days will both teams, working together, complete this job?

2) A passenger train covers the distance between two cities in 10 hours, and a freight train covers this distance in 15 hours. Both trains left these cities at the same time towards each other. In how many hours will they meet?

543. 1) A fast train covers the distance between two cities in 6 1/4 hours, and a passenger train in 7 1/2 hours. How many hours later will these trains meet if they leave both cities at the same time towards each other? (Round answer to the nearest 1 hour.)

2) Two motorcyclists left simultaneously from two cities towards each other. One motorcyclist can travel the entire distance between these cities in 6 hours, and another in 5 hours. How many hours after departure will the motorcyclists meet? (Round answer to the nearest 1 hour.)

544. 1) Three vehicles of different carrying capacity can transport some cargo, working separately: the first in 10 hours, the second in 12 hours. and the third in 15 hours. In how many hours can they transport the same cargo, working together?

2) Two trains leave two stations simultaneously towards each other: the first train covers the distance between these stations in 12 1/2 hours, and the second in 18 3/4 hours. How many hours after leaving will the trains meet?

545. 1) Two taps are connected to the bathtub. Through one of them the bath can be filled in 12 minutes, through the other 1 1/2 times faster. How many minutes will it take to fill 5/6 of the entire bathtub if you open both taps at once?

2) Two typists must retype the manuscript. The first driver can complete this work in 3 1/3 days, and the second 1 1/2 times faster. How many days will it take both typists to complete the job if they work simultaneously?

546. 1) The pool is filled with the first pipe in 5 hours, and through the second pipe it can be emptied in 6 hours. After how many hours will the entire pool be filled if both pipes are opened at the same time?

Note. In an hour, the pool is filled to (1/5 - 1/6 of its capacity.)

2) Two tractors plowed the field in 6 hours. The first tractor, working alone, could plow this field in 15 hours. How many hours would it take the second tractor, working alone, to plow this field?

547 *. Two trains leave two stations simultaneously towards each other and meet after 18 hours. after his release. How long does it take the second train to cover the distance between stations if the first train covers this distance in 1 day 21 hours?

548 *. The pool is filled with two pipes. First they opened the first pipe, and then after 3 3/4 hours, when half of the pool was filled, they opened the second pipe. After 2 1/2 hours of working together, the pool was full. Determine the capacity of the pool if 200 buckets of water per hour poured through the second pipe.

549. 1) A courier train left Leningrad for Moscow and travels 1 km in 3/4 minutes. 1/2 hour after this train left Moscow, a fast train left Moscow for Leningrad, the speed of which was equal to 3/4 the speed of the express train. At what distance will the trains be from each other 2 1/2 hours after the courier train leaves, if the distance between Moscow and Leningrad is 650 km?

2) From the collective farm to the city 24 km. A truck leaves the collective farm and travels 1 km in 2 1/2 minutes. After 15 min. After this car left the city, a cyclist drove out to the collective farm, at a speed half as fast as the speed of the truck. How long after leaving will the cyclist meet the truck?

550. 1) A pedestrian came out of one village. 4 1/2 hours after the pedestrian left, a cyclist rode in the same direction, whose speed was 2 1/2 times the speed of the pedestrian. How many hours after the pedestrian leaves will the cyclist overtake him?

2) A fast train travels 187 1/2 km in 3 hours, and a freight train travels 288 km in 6 hours. 7 1/4 hours after the freight train leaves, an ambulance departs in the same direction. How long will it take the fast train to catch up with the freight train?

551. 1) From two collective farms through which the road to the regional center passes, two collective farmers rode out to the district at the same time on horseback. The first of them traveled 8 3/4 km per hour, and the second was 1 1/7 times more than the first. The second collective farmer caught up with the first after 3 4/5 hours. Determine the distance between collective farms.

2) 26 1/3 hours after the departure of the Moscow-Vladivostok train, the average speed of which was 60 km per hour, a TU-104 plane took off in the same direction, at a speed 14 1/6 times the speed of the train. How many hours after departure will the plane catch up with the train?

552. 1) The distance between cities along the river is 264 km. The steamer covered this distance downstream in 18 hours, spending 1/12 of this time stopping. The speed of the river is 1 1/2 km per hour. How long would it take a steamship to travel 87 km without stopping in still water?

2) A motor boat traveled 207 km along the river in 13 1/2 hours, spending 1/9 of this time on stops. The speed of the river is 1 3/4 km per hour. How many kilometers can this boat travel in still water in 2 1/2 hours?

553. The boat covered a distance of 52 km across the reservoir without stopping in 3 hours 15 minutes. Further, going along the river against the current, the speed of which is 1 3/4 km per hour, this boat covered 28 1/2 km in 2 1/4 hours, making 3 stops of equal duration. How many minutes did the boat wait at each stop?

554. From Leningrad to Kronstadt at 12 o'clock. The steamer left in the afternoon and covered the entire distance between these cities in 1 1/2 hours. On the way, he met another ship that left Kronstadt for Leningrad at 12:18 p.m. and walking at 1 1/4 times the speed of the first. At what time did the two ships meet?

555. The train had to cover a distance of 630 km in 14 hours. Having covered 2/3 of this distance, he was detained for 1 hour 10 minutes. At what speed should he continue his journey in order to reach his destination without delay?

556. At 4:20 a.m. morning a freight train left Kyiv for Odessa with average speed 31 1/5 km per hour. After some time, a mail train came out of Odessa to meet him, the speed of which was 1 17/39 times higher than the speed of a freight train, and met the freight train 6 1/2 hours after its departure. At what time did the postal train leave Odessa, if the distance between Kiev and Odessa is 663 km?

557*. The clock shows noon. How long will it take for the hour and minute hands to coincide?

558. 1) The plant has three workshops. The number of workers in the first workshop is 9/20 of all workers of the plant, in the second workshop there are 1 1/2 times fewer workers than in the first, and in the third workshop there are 300 fewer workers than in the second. How many workers are there in the factory?

2) There are three secondary schools in the city. The number of students in the first school is 3/10 of all students in these three schools; in the second school there are 1 1/2 times more students than in the first, and in the third school there are 420 fewer students than in the second. How many students are there in the three schools?

559. 1) Two combine operators worked in the same area. After one combiner harvested 9/16 of the entire plot, and the second 3/8 of the same plot, it turned out that the first combiner harvested 97 1/2 hectares more than the second. On average, 32 1/2 quintals of grain were threshed from each hectare. How many centners of grain did each combine operator thresh?

2) Two brothers bought a camera. One had 5/8, and the second 4/7 of the cost of the camera, and the first had 2 rubles worth. 25 kopecks more than the second one. Everyone paid half the cost of the device. How much money does everyone have left?

560. 1) A passenger car leaves city A for city B, the distance between them is 215 km, at a speed of 50 km per hour. At the same time, a truck left city B for city A. How many kilometers did the passenger car travel before meeting the truck, if the truck's speed per hour was 18/25 the speed of the passenger car?

2) Between cities A and B 210 km. A passenger car left city A for city B. At the same time, a truck left city B for city A. How many kilometers did the truck travel before meeting the passenger car, if the passenger car was traveling at a speed of 48 km per hour, and the speed of the truck per hour was 3/4 of the speed of the passenger car?

561. The collective farm harvested wheat and rye. 20 hectares more were sown with wheat than with rye. The total rye harvest amounted to 5/6 of the total wheat harvest with a yield of 20 c per 1 ha for both wheat and rye. The collective farm sold 7/11 of the entire harvest of wheat and rye to the state, and left the rest of the grain to satisfy its needs. How many trips did the two-ton trucks need to make to remove the bread sold to the state?

562. Rye and wheat flour were brought to the bakery. The weight of wheat flour was 3/5 of the weight of rye flour, and 4 tons more rye flour was brought than wheat flour. How much wheat and how much rye bread will the bakery bake from this flour if the baked goods make up 2/5 of the total flour?

563. Within three days, a team of workers completed 3/4 of the entire work on repairing the highway between the two collective farms. On the first day, 2 2/5 km of this highway were repaired, on the second day 1 1/2 times more than on the first, and on the third day 5/8 of what was repaired in the first two days together. Find the length of the highway between collective farms.

564. Fill in the empty spaces in the table, where S is the area of ​​the rectangle, A- the base of the rectangle, a h-height (width) of the rectangle.

565. 1) The length of a rectangular plot of land is 120 m, and the width of the plot is 2/5 of its length. Find the perimeter and area of ​​the site.

2) The width of the rectangular section is 250 m, and its length is 1 1/2 times the width. Find the perimeter and area of ​​the site.

566. 1) The perimeter of the rectangle is 6 1/2 inch, its base is 1/4 inch greater than its height. Find the area of ​​this rectangle.

2) The perimeter of the rectangle is 18 cm, its height is 2 1/2 cm less than the base. Find the area of ​​the rectangle.

567. Calculate the areas of the figures shown in Figure 30 by dividing them into rectangles and finding the dimensions of the rectangle by measurement.

568. 1) How many sheets of dry plaster will be required to cover the ceiling of a room whose length is 4 1/2 m and width 4 m, if the dimensions of the plaster sheet are 2 m x l 1/2 m?

2) How many boards 4 1/2 m long and 1/4 m wide are needed to lay a floor that is 4 1/2 m long and 3 1/2 m wide?

569. 1) A rectangular plot 560 m long and 3/4 of its length wide was sown with beans. How many seeds were required to sow the plot if 1 centner was sown per 1 hectare?

2) A wheat harvest of 25 quintals per hectare was collected from a rectangular field. How much wheat was harvested from the entire field if the length of the field is 800 m and the width is 3/8 of its length?

570 . 1) A rectangular plot of land, 78 3/4 m long and 56 4/5 m wide, is built up so that 4/5 of its area is occupied by buildings. Determine the area of ​​land under the buildings.

2) On a rectangular plot of land, the length of which is 9/20 km and the width is 4/9 of its length, the collective farm plans to lay out a garden. How many trees will be planted in this garden if an average area of ​​36 sq.m. is required for each tree?

571. 1) For normal daylight illumination of the room, it is necessary that the area of ​​all windows be at least 1/5 of the floor area. Determine whether there is enough light in a room whose length is 5 1/2 m and width 4 m. Does the room have one window measuring 1 1/2 m x 2 m?

2) Using the condition of the previous problem, find out whether there is enough light in your classroom.

572. 1) The barn has dimensions of 5 1/2 m x 4 1/2 m x 2 1/2 m. How much hay (by weight) will fit in this barn if it is filled to 3/4 of its height and if 1 cu. m of hay weighs 82 kg?

2) The woodpile has the shape of a rectangular parallelepiped, the dimensions of which are 2 1/2 m x 3 1/2 m x 1 1/2 m. What is the weight of the woodpile if 1 cubic. m of firewood weighs 600 kg?

573. 1) A rectangular aquarium is filled with water up to 3/5 of its height. The length of the aquarium is 1 1/2 m, width 4/5 m, height 3/4 m. How many liters of water are poured into the aquarium?

2) A pool in the shape of a rectangular parallelepiped is 6 1/2 m long, 4 m wide and 2 m high. The pool is filled with water up to 3/4 of its height. Calculate the amount of water poured into the pool.

574. A fence needs to be built around a rectangular piece of land, 75 m long and 45 m wide. How many cubic meters of boards should go into its construction if the thickness of the board is 2 1/2 cm and the height of the fence should be 2 1/4 m?

575. 1) What is the angle between the minute hand and the hour hand at 13 o'clock? at 15 o'clock? at 17 o'clock? at 21 o'clock? at 23:30?

2) How many degrees will the hour hand rotate in 2 hours? 5 o'clock? 8 o'clock? 30 min.?

3) How many degrees does an arc equal to half a circle contain? 1/4 circle? 1/24 of a circle? 5/24 circles?

576. 1) Using a protractor, draw: a) a right angle; b) an angle of 30°; c) an angle of 60°; d) angle of 150°; e) an angle of 55°.

2) Using a protractor, measure the angles of the figure and find the sum of all the angles of each figure (Fig. 31).

577. Follow these steps:

578. 1) The semicircle is divided into two arcs, one of which is 100° larger than the other. Find the size of each arc.

2) The semicircle is divided into two arcs, one of which is 15° less than the other. Find the size of each arc.

3) The semicircle is divided into two arcs, one of which is twice as large as the other. Find the size of each arc.

4) The semicircle is divided into two arcs, one of which is 5 times smaller than the other. Find the size of each arc.

579. 1) The diagram “Population Literacy in the USSR” (Fig. 32) shows the number of literate people per hundred people of the population. Based on the data in the diagram and its scale, determine the number of literate men and women for each of the indicated years.

Write the results in the table:

2) Using the data from the diagram “Soviet envoys into space” (Fig. 33), create tasks.

580. 1) According to the pie chart “Daily routine for a fifth grade student” (Fig. 34), fill out the table and answer the questions: what part of the day is allocated to sleep? for homework? to school?

2) Construct a pie chart about your daily routine.

This article examines operations on fractions. Rules for addition, subtraction, multiplication, division or exponentiation of fractions of the form A B will be formed and justified, where A and B can be numbers, numerical expressions or expressions with variables. In conclusion, examples of solutions with detailed descriptions will be considered.

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Rules for performing operations with general numerical fractions

Numerical fractions general view have a numerator and denominator that contain natural numbers or numerical expressions. If we consider fractions such as 3 5, 2, 8 4, 1 + 2 3 4 (5 - 2), 3 4 + 7 8 2, 3 - 0, 8, 1 2 2, π 1 - 2 3 + π, 2 0, 5 ln 3, then it is clear that the numerator and denominator can have not only numbers, but also expressions of various types.

Definition 1

There are rules by which operations with ordinary fractions are carried out. It is also suitable for general fractions:

• When subtracting fractions with like denominators, only the numerators are added, and the denominator remains the same, namely: a d ± c d = a ± c d, the values ​​a, c and d ≠ 0 are some numbers or numerical expressions.
• When adding or subtracting a fraction with different denominators, it is necessary to reduce it to a common denominator, and then add or subtract the resulting fractions with the same exponents. Literally it looks like this: a b ± c d = a · p ± c · r s, where the values ​​a, b ≠ 0, c, d ≠ 0, p ≠ 0, r ≠ 0, s ≠ 0 are real numbers, and b · p = d · r = s . When p = d and r = b, then a b ± c d = a · d ± c · d b · d.
• When multiplying fractions, the action is performed with numerators, after which with denominators, then we get a b · c d = a · c b · d, where a, b ≠ 0, c, d ≠ 0 act as real numbers.
• When dividing a fraction by a fraction, we multiply the first by the second inverse, that is, we swap the numerator and denominator: a b: c d = a b · d c.

Rationale for the rules

Definition 2

There are the following mathematical points that you should rely on when calculating:

• the slash means the division sign;
• division by a number is treated as multiplication by its reciprocal value;
• application of the property of operations with real numbers;
• application of the basic property of fractions and numerical inequalities.

With their help, you can perform transformations of the form:

a d ± c d = a · d - 1 ± c · d - 1 = a ± c · d - 1 = a ± c d ; a b ± c d = a · p b · p ± c · r d · r = a · p s ± c · e s = a · p ± c · r s ; a b · c d = a · d b · d · b · c b · d = a · d · a · d - 1 · b · c · b · d - 1 = = a · d · b · c · b · d - 1 · b · d - 1 = a · d · b · c b · d · b · d - 1 = = (a · c) · (b · d) - 1 = a · c b · d

Examples

In the previous paragraph it was said about operations with fractions. It is after this that the fraction needs to be simplified. This topic was discussed in detail in the paragraph on converting fractions.

First, let's look at an example of adding and subtracting fractions with the same denominator.

Example 1

Given the fractions 8 2, 7 and 1 2, 7, then according to the rule it is necessary to add the numerator and rewrite the denominator.

Solution

Then we get a fraction of the form 8 + 1 2, 7. After performing the addition, we obtain a fraction of the form 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3. So, 8 2, 7 + 1 2, 7 = 8 + 1 2, 7 = 9 2, 7 = 90 27 = 3 1 3.

Answer: 8 2 , 7 + 1 2 , 7 = 3 1 3

There is another solution. To begin with, we switch to the form of an ordinary fraction, after which we perform a simplification. It looks like this:

8 2 , 7 + 1 2 , 7 = 80 27 + 10 27 = 90 27 = 3 1 3

Example 2

Let's subtract from 1 - 2 3 · log 2 3 · log 2 5 + 1 a fraction of the form 2 3 3 · log 2 3 · log 2 5 + 1 .

Since equal denominators are given, it means that we are calculating a fraction with the same denominator. We get that

1 - 2 3 log 2 3 log 2 5 + 1 - 2 3 3 log 2 3 log 2 5 + 1 = 1 - 2 - 2 3 3 log 2 3 log 2 5 + 1

There are examples of calculating fractions with different denominators. An important point is reduction to a common denominator. Without this, we will not be able to perform further operations with fractions.

The process is vaguely reminiscent of reduction to a common denominator. That is, the least common divisor in the denominator is searched for, after which the missing factors are added to the fractions.

If the fractions being added do not have common factors, then their product can become one.

Example 3

Let's look at the example of adding fractions 2 3 5 + 1 and 1 2.

Solution

In this case, the common denominator is the product of the denominators. Then we get that 2 · 3 5 + 1. Then, when setting additional factors, we have that for the first fraction it is equal to 2, and for the second it is 3 5 + 1. After multiplication, the fractions are reduced to the form 4 2 · 3 5 + 1. The general reduction of 1 2 will be 3 5 + 1 2 · 3 5 + 1. We add the resulting fractional expressions and get that

2 3 5 + 1 + 1 2 = 2 2 2 3 5 + 1 + 1 3 5 + 1 2 3 5 + 1 = = 4 2 3 5 + 1 + 3 5 + 1 2 3 5 + 1 = 4 + 3 5 + 1 2 3 5 + 1 = 5 + 3 5 2 3 5 + 1

Answer: 2 3 5 + 1 + 1 2 = 5 + 3 5 2 3 5 + 1

When we are dealing with general fractions, then we usually do not talk about the lowest common denominator. It is unprofitable to take the product of the numerators as the denominator. First you need to check if there is a number that is less in value than their product.

Example 4

Let's consider the example of 1 6 · 2 1 5 and 1 4 · 2 3 5, when their product is equal to 6 · 2 1 5 · 4 · 2 3 5 = 24 · 2 4 5. Then we take 12 · 2 3 5 as the common denominator.

Let's look at examples of multiplying general fractions.

Example 5

To do this, you need to multiply 2 + 1 6 and 2 · 5 3 · 2 + 1.

Solution

Following the rule, it is necessary to rewrite and write the product of the numerators as a denominator. We get that 2 + 1 6 2 5 3 2 + 1 2 + 1 2 5 6 3 2 + 1. Once a fraction has been multiplied, you can make reductions to simplify it. Then 5 · 3 3 2 + 1: 10 9 3 = 5 · 3 3 2 + 1 · 9 3 10.

Using the rule for transition from division to multiplication by a reciprocal fraction, we obtain a fraction that is the reciprocal of the given one. To do this, the numerator and denominator are swapped. Let's look at an example:

5 3 3 2 + 1: 10 9 3 = 5 3 3 2 + 1 9 3 10

Then they must multiply and simplify the resulting fraction. If necessary, get rid of irrationality in the denominator. We get that

5 3 3 2 + 1: 10 9 3 = 5 3 3 9 3 10 2 + 1 = 5 2 10 2 + 1 = 3 2 2 + 1 = 3 2 - 1 2 2 + 1 2 - 1 = 3 2 - 1 2 2 2 - 1 2 = 3 2 - 1 2

Answer: 5 3 3 2 + 1: 10 9 3 = 3 2 - 1 2

This paragraph is applicable when the number or numeric expression can be presented as a fraction with a denominator equal to 1, then the action with such a fraction is considered as a separate paragraph. For example, the expression 1 6 · 7 4 - 1 · 3 shows that the root of 3 can be replaced by another 3 1 expression. Then this entry will look like multiplying two fractions of the form 1 6 · 7 4 - 1 · 3 = 1 6 · 7 4 - 1 · 3 1.

Performing Operations on Fractions Containing Variables

The rules discussed in the first article are applicable to operations with fractions containing variables. Consider the subtraction rule when the denominators are the same.

It is necessary to prove that A, C and D (D not equal to zero) can be any expressions, and the equality A D ± C D = A ± C D is equivalent to its range of permissible values.

It is necessary to take a set of ODZ variables. Then A, C, D must take the corresponding values ​​a 0 , c 0 and d 0. Substitution of the form A D ± C D results in a difference of the form a 0 d 0 ± c 0 d 0 , where, using the addition rule, we obtain a formula of the form a 0 ± c 0 d 0 . If we substitute the expression A ± C D, then we get the same fraction of the form a 0 ± c 0 d 0. From here we conclude that the selected value that satisfies the ODZ, A ± C D and A D ± C D are considered equal.

For any value of the variables, these expressions will be equal, that is, they are called identically equal. This means that this expression is considered a provable equality of the form A D ± C D = A ± C D .

Examples of adding and subtracting fractions with variables

When available same denominators, you only need to add or subtract the numerators. This fraction can be simplified. Sometimes you have to work with fractions that are identically equal, but at first glance this is not noticeable, since some transformations must be performed. For example, x 2 3 x 1 3 + 1 and x 1 3 + 1 2 or 1 2 sin 2 α and sin a cos a. Most often, a simplification of the original expression is required in order to see the same denominators.

Example 6

Calculate: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) , x - 1 x - 1 + x x + 1 .

Solution

1. To make the calculation, you need to subtract fractions that have the same denominator. Then we get that x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + 1 - 5 - x x + x - 2 . After which you can expand the brackets and add similar terms. We get that x 2 + 1 - 5 - x x + x - 2 = x 2 + 1 - 5 + x x + x - 2 = x 2 + x - 4 x + x - 2
2. Since the denominators are the same, all that remains is to add the numerators, leaving the denominator: l g 2 x + 4 x (l g x + 2) + 4 l g x x (l g x + 2) = l g 2 x + 4 + 4 x (l g x + 2)
   The addition has been completed. It can be seen that it is possible to reduce the fraction. Its numerator can be folded using the formula for the square of the sum, then we get (l g x + 2) 2 from abbreviated multiplication formulas. Then we get that
   l g 2 x + 4 + 2 l g x x (l g x + 2) = (l g x + 2) 2 x (l g x + 2) = l g x + 2 x
3. Given fractions of the form x - 1 x - 1 + x x + 1 with different denominators. After the transformation, you can move on to addition.

Let's consider a twofold solution.

The first method is that the denominator of the first fraction is factorized using squares, with its subsequent reduction. We get a fraction of the form

x - 1 x - 1 = x - 1 (x - 1) x + 1 = 1 x + 1

So x - 1 x - 1 + x x + 1 = 1 x + 1 + x x + 1 = 1 + x x + 1 .

In this case, it is necessary to get rid of irrationality in the denominator.

1 + x x + 1 = 1 + x x - 1 x + 1 x - 1 = x - 1 + x x - x x - 1

The second method is to multiply the numerator and denominator of the second fraction by the expression x - 1. Thus, we get rid of irrationality and move on to adding fractions with the same denominator. Then

x - 1 x - 1 + x x + 1 = x - 1 x - 1 + x x - 1 x + 1 x - 1 = = x - 1 x - 1 + x x - x x - 1 = x - 1 + x · x - x x - 1

Answer: 1) x 2 + 1 x + x - 2 - 5 - x x + x - 2 = x 2 + x - 4 x + x - 2, 2) l g 2 x + 4 x · (l g x + 2) + 4 · l g x x · (l g x + 2) = l g x + 2 x, 3) x - 1 x - 1 + x x + 1 = x - 1 + x · x - x x - 1 .

In the last example we found that reduction to a common denominator is inevitable. To do this, you need to simplify the fractions. When adding or subtracting, you always need to look for a common denominator, which looks like the product of the denominators with additional factors added to the numerators.

Example 7

Calculate the values ​​of the fractions: 1) x 3 + 1 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) (2 x - 4) - sin x x 5 ln (x + 1) (2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x

Solution

1. The denominator does not require any complex calculations, so you need to choose their product of the form 3 x 7 + 2 · 2, then choose x 7 + 2 · 2 for the first fraction as an additional factor, and 3 for the second. When multiplying, we get a fraction of the form x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 3 x 7 + 2 2 + 3 1 3 x 7 + 2 2 = = x x 7 + 2 2 + 3 3 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2
2. It can be seen that the denominators are presented in the form of a product, which means that additional transformations are unnecessary. The common denominator will be considered to be a product of the form x 5 · ln 2 x + 1 · 2 x - 4 . Hence x 4 is an additional factor to the first fraction, and ln(x + 1) to the second. Then we subtract and get:
   x + 1 x · ln 2 (x + 1) · 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x + 1 · x 4 x 5 · ln 2 (x + 1 ) · 2 x - 4 - sin x · ln x + 1 x 5 · ln 2 (x + 1) · (2 ​​x - 4) = = x + 1 · x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4) = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · (2 ​​x - 4 )
3. This example makes sense when working with fraction denominators. It is necessary to apply the formulas for the difference of squares and the square of the sum, since they will make it possible to move on to an expression of the form 1 cos x - x · cos x + x + 1 (cos x + x) 2. It can be seen that the fractions are reduced to a common denominator. We get that cos x - x · cos x + x 2 .

Then we get that

1 cos 2 x - x + 1 cos 2 x + 2 cos x x + x = = 1 cos x - x cos x + x + 1 cos x + x 2 = = cos x + x cos x - x cos x + x 2 + cos x - x cos x - x cos x + x 2 = = cos x + x + cos x - x cos x - x cos x + x 2 = 2 cos x cos x - x cos x + x 2

Answer:

1) x 3 + 1 x 7 + 2 2 = x x 7 + 2 2 x + 3 3 x 7 + 2 2, 2) x + 1 x ln 2 (x + 1) 2 x - 4 - sin x x 5 · ln (x + 1) · 2 x - 4 = = x · x 4 + x 4 - sin x · ln (x + 1) x 5 · ln 2 (x + 1) · ( 2 x - 4) , 3) ​​1 cos 2 x - x + 1 cos 2 x + 2 · cos x · x + x = 2 · cos x cos x - x · cos x + x 2 .

Examples of multiplying fractions with variables

When multiplying fractions, the numerator is multiplied by the numerator and the denominator by the denominator. Then you can apply the reduction property.

Example 8

Multiply the fractions x + 2 · x x 2 · ln x 2 · ln x + 1 and 3 · x 2 1 3 · x + 1 - 2 sin 2 · x - x.

Solution

Multiplication needs to be done. We get that

x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = = x - 2 x 3 x 2 1 3 x + 1 - 2 x 2 ln x 2 ln x + 1 sin (2 x - x)

The number 3 is moved to the first place for the convenience of calculations, and you can reduce the fraction by x 2, then we get an expression of the form

3 x - 2 x x 1 3 x + 1 - 2 ln x 2 ln x + 1 sin (2 x - x)

Answer: x + 2 x x 2 ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x) = 3 x - 2 x x 1 3 x + 1 - 2 ln x 2 · ln x + 1 · sin (2 · x - x) .

Division

Division of fractions is similar to multiplication, since the first fraction is multiplied by the second reciprocal. If we take for example the fraction x + 2 x x 2 ln x 2 ln x + 1 and divide by 3 x 2 1 3 x + 1 - 2 sin 2 x - x, then it can be written as

x + 2 · x x 2 · ln x 2 · ln x + 1: 3 · x 2 1 3 · x + 1 - 2 sin (2 · x - x) , then replace with a product of the form x + 2 · x x 2 · ln x 2 ln x + 1 3 x 2 1 3 x + 1 - 2 sin (2 x - x)

Exponentiation

Let's move on to considering operations with general fractions with exponentiation. If there is a power with a natural exponent, then the action is considered as multiplication of equal fractions. But it is recommended to use a general approach based on the properties of degrees. Any expressions A and C, where C is not identically equal to zero, and any real r on the ODZ for an expression of the form A C r the equality A C r = A r C r is valid. The result is a fraction raised to a power. For example, consider:

x 0, 7 - π · ln 3 x - 2 - 5 x + 1 2, 5 = = x 0, 7 - π · ln 3 x - 2 - 5 2, 5 x + 1 2, 5

Procedure for performing operations with fractions

Operations on fractions are performed according to certain rules. In practice, we notice that an expression may contain several fractions or fractional expressions. Then it is necessary to perform all actions in strict order: raise to a power, multiply, divide, then add and subtract. If there are parentheses, the first action is performed in them.

Example 9

Calculate 1 - x cos x - 1 c o s x · 1 + 1 x .

Solution

Since we have the same denominator, then 1 - x cos x and 1 c o s x, but subtractions cannot be performed according to the rule; first, the actions in parentheses are performed, then multiplication, and then addition. Then when calculating we get that

1 + 1 x = 1 1 + 1 x = x x + 1 x = x + 1 x

When substituting the expression into the original one, we get that 1 - x cos x - 1 cos x · x + 1 x. When multiplying fractions we have: 1 cos x · x + 1 x = x + 1 cos x · x. Having made all the substitutions, we get 1 - x cos x - x + 1 cos x · x. Now you need to work with fractions that have different denominators. We get:

x · 1 - x cos x · x - x + 1 cos x · x = x · 1 - x - 1 + x cos x · x = = x - x - x - 1 cos x · x = - x + 1 cos x x

Answer: 1 - x cos x - 1 c o s x · 1 + 1 x = - x + 1 cos x · x .

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Lesson content

Adding fractions with like denominators

There are two types of addition of fractions:

1. Adding fractions with like denominators
2. Adding fractions with different denominators

First, let's learn the addition of fractions with like denominators. Everything is simple here. To add fractions with the same denominators, you need to add their numerators and leave the denominator unchanged. For example, let's add the fractions and . Add the numerators and leave the denominator unchanged:

This example can be easily understood if we remember the pizza, which is divided into four parts. If you add pizza to pizza, you get pizza:

Example 2. Add fractions and .

The answer turned out to be an improper fraction. When the end of the task comes, it is customary to get rid of improper fractions. To get rid of an improper fraction, you need to select the whole part of it. In our case, the whole part is easily isolated - two divided by two equals one:

This example can be easily understood if we remember about a pizza that is divided into two parts. If you add more pizza to the pizza, you get one whole pizza:

Example 3. Add fractions and .

Again, we add up the numerators and leave the denominator unchanged:

This example can be easily understood if we remember the pizza, which is divided into three parts. If you add more pizza to the pizza, you get pizza:

Example 4. Find the value of an expression

This example is solved in exactly the same way as the previous ones. The numerators must be added and the denominator left unchanged:

Let's try to depict our solution using a drawing. If you add pizzas to a pizza and add more pizzas, you get 1 whole pizza and more pizzas.

As you can see, there is nothing complicated about adding fractions with the same denominators. It is enough to understand the following rules:

1. To add fractions with the same denominator, you need to add their numerators and leave the denominator unchanged;

Adding fractions with different denominators

Now let's learn how to add fractions with different denominators. When adding fractions, the denominators of the fractions must be the same. But they are not always the same.

For example, fractions can be added because they have the same denominators.

But fractions cannot be added right away, since these fractions have different denominators. In such cases, fractions must be reduced to the same (common) denominator.

There are several ways to reduce fractions to the same denominator. Today we will look at only one of them, since the other methods may seem complicated for a beginner.

The essence of this method is that first the LCM of the denominators of both fractions is searched. The LCM is then divided by the denominator of the first fraction to obtain the first additional factor. They do the same with the second fraction - the LCM is divided by the denominator of the second fraction and a second additional factor is obtained.

The numerators and denominators of the fractions are then multiplied by their additional factors. As a result of these actions, fractions that had different denominators turn into fractions that have the same denominators. And we already know how to add such fractions.

Example 1. Let's add the fractions and

First of all, we find the least common multiple of the denominators of both fractions. The denominator of the first fraction is the number 3, and the denominator of the second fraction is the number 2. The least common multiple of these numbers is 6

LCM (2 and 3) = 6

Now let's return to fractions and . First, divide the LCM by the denominator of the first fraction and get the first additional factor. LCM is the number 6, and the denominator of the first fraction is the number 3. Divide 6 by 3, we get 2.

The resulting number 2 is the first additional multiplier. We write it down to the first fraction. To do this, make a small oblique line over the fraction and write down the additional factor found above it:

We do the same with the second fraction. We divide the LCM by the denominator of the second fraction and get the second additional factor. LCM is the number 6, and the denominator of the second fraction is the number 2. Divide 6 by 2, we get 3.

The resulting number 3 is the second additional multiplier. We write it down to the second fraction. Again, we make a small oblique line over the second fraction and write down the additional factor found above it:

Now we have everything ready for addition. It remains to multiply the numerators and denominators of the fractions by their additional factors:

Look carefully at what we have come to. We came to the conclusion that fractions that had different denominators turned into fractions that had the same denominators. And we already know how to add such fractions. Let's take this example to the end:

This completes the example. It turns out to add .

Let's try to depict our solution using a drawing. If you add pizza to a pizza, you get one whole pizza and another sixth of a pizza:

Reducing fractions to the same (common) denominator can also be depicted using a picture. Reducing the fractions and to a common denominator, we got the fractions and . These two fractions will be represented by the same pieces of pizza. The only difference will be that this time they will be divided into equal shares (reduced to the same denominator).

The first drawing represents a fraction (four pieces out of six), and the second drawing represents a fraction (three pieces out of six). Adding these pieces we get (seven pieces out of six). This fraction is improper, so we highlighted the whole part of it. As a result, we got (one whole pizza and another sixth pizza).

Please note that we have described this example in too much detail. IN educational institutions It’s not customary to write in such detail. You need to be able to quickly find the LCM of both denominators and additional factors to them, as well as quickly multiply the found additional factors by your numerators and denominators. If we were at school, we would have to write this example as follows:

But there is also another side to the coin. If you do not take detailed notes in the first stages of studying mathematics, then questions of the sort begin to appear. “Where does that number come from?”, “Why do fractions suddenly turn into completely different fractions? «.

To make it easier to add fractions with different denominators, you can use the following step-by-step instructions:

1. Find the LCM of the denominators of fractions;
2. Divide the LCM by the denominator of each fraction and obtain an additional factor for each fraction;
3. Multiply the numerators and denominators of fractions by their additional factors;
4. Add fractions that have the same denominators;
5. If the answer turns out to be an improper fraction, then select its whole part;

Example 2. Find the value of an expression .

Let's use the instructions given above.

Step 1. Find the LCM of the denominators of the fractions

Find the LCM of the denominators of both fractions. The denominators of fractions are the numbers 2, 3 and 4

Step 2. Divide the LCM by the denominator of each fraction and get an additional factor for each fraction

Divide the LCM by the denominator of the first fraction. LCM is the number 12, and the denominator of the first fraction is the number 2. Divide 12 by 2, we get 6. We got the first additional factor 6. We write it above the first fraction:

Now we divide the LCM by the denominator of the second fraction. LCM is the number 12, and the denominator of the second fraction is the number 3. Divide 12 by 3, we get 4. We get the second additional factor 4. We write it above the second fraction:

Now we divide the LCM by the denominator of the third fraction. LCM is the number 12, and the denominator of the third fraction is the number 4. Divide 12 by 4, we get 3. We get the third additional factor 3. We write it above the third fraction:

Step 3. Multiply the numerators and denominators of the fractions by their additional factors

We multiply the numerators and denominators by their additional factors:

Step 4. Add fractions with the same denominators

We came to the conclusion that fractions that had different denominators turned into fractions that had the same (common) denominators. All that remains is to add these fractions. Add it up:

The addition didn't fit on one line, so we moved the remaining expression to the next line. This is allowed in mathematics. When an expression does not fit on one line, it is moved to the next line, and it is necessary to put an equal sign (=) at the end of the first line and at the beginning of the new line. The equal sign on the second line indicates that this is a continuation of the expression that was on the first line.

Step 5. If the answer turns out to be an improper fraction, then select the whole part of it

Our answer turned out to be an improper fraction. We have to highlight a whole part of it. We highlight:

We received an answer

Subtracting fractions with like denominators

There are two types of subtraction of fractions:

1. Subtracting fractions with like denominators
2. Subtracting fractions with different denominators

First, let's learn how to subtract fractions with like denominators. Everything is simple here. To subtract another from one fraction, you need to subtract the numerator of the second fraction from the numerator of the first fraction, but leave the denominator the same.

For example, let's find the value of the expression . To solve this example, you need to subtract the numerator of the second fraction from the numerator of the first fraction, and leave the denominator unchanged. Let's do this:

This example can be easily understood if we remember the pizza, which is divided into four parts. If you cut pizzas from a pizza, you get pizzas:

Example 2. Find the value of the expression.

Again, from the numerator of the first fraction, subtract the numerator of the second fraction, and leave the denominator unchanged:

This example can be easily understood if we remember the pizza, which is divided into three parts. If you cut pizzas from a pizza, you get pizzas:

Example 3. Find the value of an expression

This example is solved in exactly the same way as the previous ones. From the numerator of the first fraction you need to subtract the numerators of the remaining fractions:

As you can see, there is nothing complicated about subtracting fractions with the same denominators. It is enough to understand the following rules:

1. To subtract another from one fraction, you need to subtract the numerator of the second fraction from the numerator of the first fraction, and leave the denominator unchanged;
2. If the answer turns out to be an improper fraction, then you need to highlight the whole part of it.

Subtracting fractions with different denominators

For example, you can subtract a fraction from a fraction because the fractions have the same denominators. But you cannot subtract a fraction from a fraction, since these fractions have different denominators. In such cases, fractions must be reduced to the same (common) denominator.

The common denominator is found using the same principle that we used when adding fractions with different denominators. First of all, find the LCM of the denominators of both fractions. Then the LCM is divided by the denominator of the first fraction and the first additional factor is obtained, which is written above the first fraction. Similarly, the LCM is divided by the denominator of the second fraction and a second additional factor is obtained, which is written above the second fraction.

The fractions are then multiplied by their additional factors. As a result of these operations, fractions that had different denominators are converted into fractions that have the same denominators. And we already know how to subtract such fractions.

Example 1. Find the meaning of the expression:

These fractions have different denominators, so you need to reduce them to the same (common) denominator.

First we find the LCM of the denominators of both fractions. The denominator of the first fraction is the number 3, and the denominator of the second fraction is the number 4. The least common multiple of these numbers is 12

LCM (3 and 4) = 12

Now let's return to fractions and

Let's find an additional factor for the first fraction. To do this, divide the LCM by the denominator of the first fraction. LCM is the number 12, and the denominator of the first fraction is the number 3. Divide 12 by 3, we get 4. Write a four above the first fraction:

We do the same with the second fraction. Divide the LCM by the denominator of the second fraction. LCM is the number 12, and the denominator of the second fraction is the number 4. Divide 12 by 4, we get 3. Write a three over the second fraction:

Now we are ready for subtraction. It remains to multiply the fractions by their additional factors:

We came to the conclusion that fractions that had different denominators turned into fractions that had the same denominators. And we already know how to subtract such fractions. Let's take this example to the end:

We received an answer

Let's try to depict our solution using a drawing. If you cut pizza from a pizza, you get pizza

This is the detailed version of the solution. If we were at school, we would have to solve this example shorter. Such a solution would look like this:

Reducing fractions to a common denominator can also be depicted using a picture. Reducing these fractions to a common denominator, we got the fractions and . These fractions will be represented by the same pizza slices, but this time they will be divided into equal shares (reduced to the same denominator):

The first picture shows a fraction (eight pieces out of twelve), and the second picture shows a fraction (three pieces out of twelve). By cutting three pieces from eight pieces, we get five pieces out of twelve. The fraction describes these five pieces.

Example 2. Find the value of an expression

These fractions have different denominators, so first you need to reduce them to the same (common) denominator.

Let's find the LCM of the denominators of these fractions.

The denominators of the fractions are the numbers 10, 3 and 5. The least common multiple of these numbers is 30

LCM(10, 3, 5) = 30

Now we find additional factors for each fraction. To do this, divide the LCM by the denominator of each fraction.

Let's find an additional factor for the first fraction. LCM is the number 30, and the denominator of the first fraction is the number 10. Divide 30 by 10, we get the first additional factor 3. We write it above the first fraction:

Now we find an additional factor for the second fraction. Divide the LCM by the denominator of the second fraction. LCM is the number 30, and the denominator of the second fraction is the number 3. Divide 30 by 3, we get the second additional factor 10. We write it above the second fraction:

Now we find an additional factor for the third fraction. Divide the LCM by the denominator of the third fraction. LCM is the number 30, and the denominator of the third fraction is the number 5. Divide 30 by 5, we get the third additional factor 6. We write it above the third fraction:

Now everything is ready for subtraction. It remains to multiply the fractions by their additional factors:

We came to the conclusion that fractions that had different denominators turned into fractions that had the same (common) denominators. And we already know how to subtract such fractions. Let's finish this example.

The continuation of the example will not fit on one line, so we move the continuation to the next line. Don't forget about the equal sign (=) on the new line:

The answer turned out to be a regular fraction, and everything seems to suit us, but it is too cumbersome and ugly. We should make it simpler. What can be done? You can shorten this fraction.

To reduce a fraction, you need to divide its numerator and denominator by (GCD) of the numbers 20 and 30.

So, we find the gcd of numbers 20 and 30:

Now we return to our example and divide the numerator and denominator of the fraction by the found gcd, that is, by 10

We received an answer

Multiplying a fraction by a number

To multiply a fraction by a number, you need to multiply the numerator of the fraction by that number and leave the denominator unchanged.

Example 1. Multiply a fraction by the number 1.

Multiply the numerator of the fraction by the number 1

The recording can be understood as taking half 1 time. For example, if you take pizza once, you get pizza

From the laws of multiplication we know that if the multiplicand and the factor are swapped, the product will not change. If the expression is written as , then the product will still be equal to . Again, the rule for multiplying a whole number and a fraction works:

This notation can be understood as taking half of one. For example, if there is 1 whole pizza and we take half of it, then we will have pizza:

Example 2. Find the value of an expression

Multiply the numerator of the fraction by 4

The answer was an improper fraction. Let's highlight the whole part of it:

The expression can be understood as taking two quarters 4 times. For example, if you take 4 pizzas, you will get two whole pizzas

And if we swap the multiplicand and the multiplier, we get the expression . It will also be equal to 2. This expression can be understood as taking two pizzas from four whole pizzas:

The number being multiplied by the fraction and the denominator of the fraction are resolved if they have a common factor greater than one.

For example, an expression can be evaluated in two ways.

First way. Multiply the number 4 by the numerator of the fraction, and leave the denominator of the fraction unchanged:

Second way. The four being multiplied and the four in the denominator of the fraction can be reduced. These fours can be reduced by 4, since the greatest common divisor for two fours is the four itself:

We got the same result 3. After reducing the fours, new numbers are formed in their place: two ones. But multiplying one with three, and then dividing by one does not change anything. Therefore, the solution can be written briefly:

The reduction can be performed even when we decided to use the first method, but at the stage of multiplying the number 4 and the numerator 3 we decided to use the reduction:

But for example, the expression can only be calculated in the first way - multiply 7 by the denominator of the fraction, and leave the denominator unchanged:

This is due to the fact that the number 7 and the denominator of the fraction do not have a common divisor greater than one, and accordingly do not cancel.

Some students mistakenly shorten the number being multiplied and the numerator of the fraction. You can't do this. For example, the following entry is not correct:

Reducing a fraction means that both numerator and denominator will be divided by the same number. In the situation with the expression, division is performed only in the numerator, since writing this is the same as writing . We see that division is performed only in the numerator, and no division occurs in the denominator.

Multiplying fractions

To multiply fractions, you need to multiply their numerators and denominators. If the answer turns out to be an improper fraction, you need to highlight the whole part of it.

Example 1. Find the value of the expression.

We received an answer. It is advisable to reduce this fraction. The fraction can be reduced by 2. Then the final solution will take the following form:

The expression can be understood as taking a pizza from half a pizza. Let's say we have half a pizza:

How to take two thirds from this half? First you need to divide this half into three equal parts:

And take two from these three pieces:

We'll make pizza. Remember what pizza looks like when divided into three parts:

One piece of this pizza and the two pieces we took will have the same dimensions:

In other words, we are talking about the same size pizza. Therefore the value of the expression is

Example 2. Find the value of an expression

Multiply the numerator of the first fraction by the numerator of the second fraction, and the denominator of the first fraction by the denominator of the second fraction:

The answer was an improper fraction. Let's highlight the whole part of it:

Example 3. Find the value of an expression

Multiply the numerator of the first fraction by the numerator of the second fraction, and the denominator of the first fraction by the denominator of the second fraction:

The answer turned out to be a regular fraction, but it would be good if it was shortened. To reduce this fraction, you need to divide the numerator and denominator of this fraction by the greatest common divisor (GCD) of the numbers 105 and 450.

So, let’s find the gcd of numbers 105 and 450:

Now we divide the numerator and denominator of our answer by the gcd that we have now found, that is, by 15

Representing a whole number as a fraction

Any whole number can be represented as a fraction. For example, the number 5 can be represented as . This will not change the meaning of five, since the expression means “the number five divided by one,” and this, as we know, is equal to five:

Reciprocal numbers

Now we will get acquainted with very interesting topic in mathematics. It's called "reverse numbers".

Definition. Reverse to numbera is a number that, when multiplied bya gives one.

Let's substitute in this definition instead of the variable a number 5 and try to read the definition:

Reverse to number 5 is a number that, when multiplied by 5 gives one.

Is it possible to find a number that, when multiplied by 5, gives one? It turns out it is possible. Let's imagine five as a fraction:

Then multiply this fraction by itself, just swap the numerator and denominator. In other words, let’s multiply the fraction by itself, only upside down:

What will happen as a result of this? If we continue to solve this example, we get one:

This means that the inverse of the number 5 is the number , since when you multiply 5 by you get one.

The reciprocal of a number can also be found for any other integer.

You can also find the reciprocal of any other fraction. To do this, just turn it over.

Dividing a fraction by a number

Let's say we have half a pizza:

Let's divide it equally between two. How much pizza will each person get?

It can be seen that after dividing half the pizza, two equal pieces were obtained, each of which constitutes a pizza. So everyone gets a pizza.

1º. Integers - These are numbers used in counting. The set of all natural numbers is denoted by N, i.e. N=(1, 2, 3, …).

Fraction is a number consisting of several fractions of a unit. Common fraction is a number of the form where is a natural number n shows how many equal parts a unit is divided into, and a natural number m shows how many such equal parts are taken. Numbers m And n are called accordingly numerator And denominator fractions

If the numerator is less than the denominator, then the fraction is called correct; if the numerator is equal to or greater than the denominator, then the fraction is called wrong. A number consisting of an integer and fractional parts, called mixed number.

For example,
- correct common fractions,
- improper ordinary fractions, 1 is a mixed number.

2º. When performing operations with ordinary fractions, you should remember the following rules:

1)The main property of a fraction. If the numerator and denominator of a fraction are multiplied or divided by the same natural number, you get a fraction equal to the given one.

For example, a)
; b)
.

Dividing the numerator and denominator of a fraction by their common divisor other than one is called reducing a fraction.

2) To represent a mixed number as an improper fraction, you need to multiply its whole part by the denominator of the fractional part and add the numerator of the fractional part to the resulting product, write the resulting amount as the numerator of the fraction, and leave the denominator the same.

Similarly, any natural number can be written as an improper fraction with any denominator.

For example, a)
, because
; b)
etc.

3) To write an improper fraction as a mixed number (i.e., separate an integer part from an improper fraction), you need to divide the numerator by the denominator, take the quotient of the division as an integer part, the remainder as the numerator, and leave the denominator the same.

For example, a)
, since 200: 7 = 28 (remaining 4); b)
, since 20: 5 = 4 (remaining 0).

4) To reduce fractions to the lowest common denominator, you need to find the least common multiple (LCM) of the denominators of these fractions (it will be their lowest common denominator), divide the lowest common denominator by the denominators of these fractions (i.e. find additional factors for the fractions) , multiply the numerator and denominator of each fraction by its additional factor.

For example, let's give the fractions
to the lowest common denominator:

,
,
;

630: 18 = 35, 630: 10 = 63, 630: 21 = 30.

Means,
;
;
.

5) Rules for arithmetic operations on ordinary fractions:

a) Addition and subtraction of fractions with the same denominators is performed according to the rule:

.

b) Addition and subtraction of fractions with different denominators is carried out according to rule a), after first reducing the fractions to the lowest common denominator.

c) When adding and subtracting mixed numbers, you can turn them into improper fractions, and then perform actions according to rules a) and b),

d) When multiplying fractions, use the following rule:

.

e) To divide one fraction by another, you need to multiply the dividend by the reciprocal of the divisor:

.

f) When multiplying and dividing mixed numbers, they are first converted to improper fractions, and then rules d) and e) are used.

3º. When solving examples for all operations with fractions, remember that the operations in parentheses are performed first. Both inside and outside parentheses, multiplication and division are performed first, followed by addition and subtraction.

Let's look at the implementation of the above rules using an example.

Example 1. Calculate:
.

1)
;

2)
;

5)
. Answer: 3.