EntranceWhat an odz at the root. Range of permissible values - ODZ
Scientific adviser:
1. Introduction 3
2. Historical sketch 4
3. “Place” of ODZ when solving equations and inequalities 5-6
4. Features and dangers of ODZ 7
5. ODZ – there is a solution 8-9
6. Finding ODZ is extra work. Equivalence of transitions 10-14
7. ODZ in the Unified State Exam 15-16
8. Conclusion 17
9. Literature 18
1. Introduction
Problem: equations and inequalities in which it is necessary to find ODZ have not found a place in the algebra course for systematic presentation, which is probably why my peers and I often make mistakes when solving such examples, spending a lot of time solving them, while forgetting about ODZ.
Target: be able to analyze the situation and draw logically correct conclusions in examples where it is necessary to take into account DL.
Tasks:
1. Study theoretical material;
2. Solve many equations, inequalities: a) fractional-rational; b) irrational; c) logarithmic; d) containing inverse trigonometric functions;
3. Apply the studied materials in a situation that differs from the standard one;
4. Create a work on the topic “Area of acceptable values: theory and practice”
Project work: I started working on the project by repeating the functions I knew. The scope of many of them is limited.
ODZ occurs:
1. When deciding fractional rational equations and inequalities
2. When solving irrational equations and inequalities
3. When solving logarithmic equations and inequalities
4. When solving equations and inequalities containing inverse trigonometric functions
Having solved many examples from various sources (USE textbooks, textbooks, reference books), I systematized the solution of examples according to the following principles:
· you can solve the example and take into account the ODZ (the most common method)
· it is possible to solve the example without taking into account the ODZ
· it is only possible to come to the right decision by taking into account the ODZ.
Methods used in the work: 1) analysis; 2) statistical analysis; 3) deduction; 4) classification; 5) forecasting.
Studied the analysis Unified State Exam results over the past years. Many mistakes were made in examples in which it is necessary to take into account DL. This once again emphasizes relevance my topic.
2. Historical sketch
Like other concepts of mathematics, the concept of a function did not develop immediately, but went through a long path of development. In the work of P. Fermat “Introduction and study of plane and solid places” (1636, published 1679) it is said: “Whenever there are two unknown quantities in the final equation, there is a place.” Essentially, we are talking about functional dependence and its graphic representation(“place” in Fermat means line). The study of lines according to their equations in R. Descartes' "Geometry" (1637) also indicates a clear understanding of the mutual dependence of two variables. I. Barrow (Lectures on Geometry, 1670) establishes in geometric form the mutual inverse nature of the actions of differentiation and integration (of course, without using these terms themselves). This already indicates a completely clear mastery of the concept of function. We also find this concept in geometric and mechanical form in I. Newton. However, the term “function” first appears only in 1692 with G. Leibniz and, moreover, not quite in its modern understanding. G. Leibniz calls various segments associated with a curve (for example, the abscissa of its points) a function. In the first printed course, “Analysis of infinitesimals for the knowledge of curved lines” by L'Hopital (1696), the term “function” is not used.
The first definition of a function in a sense close to the modern one is found in I. Bernoulli (1718): “A function is a quantity composed of a variable and a constant.” This not entirely clear definition is based on the idea of specifying a function analytical formula. The same idea appears in the definition of L. Euler, given by him in “Introduction to the Analysis of Infinites” (1748): “The function of a variable quantity is an analytical expression composed in some way from this variable quantity and numbers or constant quantities.” However, L. Euler is no longer alien to the modern understanding of function, which does not connect the concept of a function with any of its analytical expressions. His “Differential Calculus” (1755) says: “When certain quantities depend on others in such a way that when the latter change they themselves are subject to change, then the former are called functions of the latter.”
WITH early XIX centuries, more and more often they define the concept of a function without mentioning its analytical representation. In the “Treatise on Differential and Integral Calculus” (1797-1802) S. Lacroix says: “Every quantity whose value depends on one or many other quantities is called a function of these latter.” In the “Analytical Theory of Heat” by J. Fourier (1822) there is a phrase: “Function f(x) denotes a completely arbitrary function, that is, a sequence of given values, whether or not subject to a general law and corresponding to all values x contained between 0 and some value x" The definition of N. I. Lobachevsky is close to modern: “... General concept function requires that the function from x name the number that is given for each x and together with x gradually changes. The value of a function can be given either by an analytical expression, or by a condition that provides a means of testing all the numbers and choosing one of them, or, finally, the dependence can exist and remain unknown. It is also said there a little lower: “The broad view of the theory allows for the existence of dependence only in the sense that numbers one with another in connection are understood as if given together.” Thus, modern definition function, free from references to the analytical task, usually attributed to P. Dirichlet (1837), was repeatedly proposed before him.
The domain of definition (admissible values) of a function y is the set of values of the independent variable x for which this function is defined, i.e., the domain of change of the independent variable (argument).
3. “Place” of the range of acceptable values when solving equations and inequalities
1. When solving fractional rational equations and inequalities the denominator must not be zero.
2. Solving irrational equations and inequalities.
2.1..gif" width="212" height="51"> .
In this case, there is no need to find the ODZ: from the first equation it follows that the obtained values of x satisfy the following inequality: https://pandia.ru/text/78/083/images/image004_33.gif" width="107" height="27 src="> is the system:
Since they enter into the equation equally, then instead of inequality, you can include inequality https://pandia.ru/text/78/083/images/image009_18.gif" width="220" height="49"> 
https://pandia.ru/text/78/083/images/image014_11.gif" width="239" height="51"> 
3. Solving logarithmic equations and inequalities.
3.1. Scheme for solving a logarithmic equation
But it is enough to check only one condition of the ODZ.
3.2..gif" width="115" height="48 src=">.gif" width="115" height="48 src=">
4. Trigonometric equations of the form are equivalent to the system (instead of inequality, you can include inequality in the system https://pandia.ru/text/78/083/images/image024_5.gif" width="377" height="23"> are equivalent to the equation
4. Features and dangers of the range of permissible values
In mathematics lessons, we are required to find the DL in each example. At the same time, according to the mathematical essence of the matter, finding the ODZ is not at all mandatory, often not necessary, and sometimes impossible - and all this without any damage to the solution of the example. On the other hand, it often happens that after solving an example, schoolchildren forget to take into account the DL, write it down as the final answer, and take into account only some conditions. This circumstance is well known, but the “war” continues every year and, it seems, will continue for a long time.
Consider, for example, the following inequality:
Here, the ODZ is sought and the inequality is solved. However, when solving this inequality, schoolchildren sometimes believe that it is quite possible to do without searching for DL, or more precisely, it is possible to do without the condition
In fact, to obtain the correct answer it is necessary to take into account both the inequality , and .
But, for example, the solution to the equation: https://pandia.ru/text/78/083/images/image032_4.gif" width="79 height=75" height="75">
which is equivalent to working with ODZ. However, in this example, such work is unnecessary - it is enough to check the fulfillment of only two of these inequalities, and any two.
Let me remind you that any equation (inequality) can be reduced to the form . ODZ is simply the domain of definition of the function on the left side. The fact that this area must be monitored follows from the definition of the root as a number from the domain of definition of a given function, thereby from the ODZ. Here is a funny example on this topic..gif" width="20" height="21 src="> has a domain of definition of a set of positive numbers (this, of course, is an agreement to consider a function with, but reasonable), and then -1 is not is the root.
5. Range of acceptable values – there is a solution
And finally, in a lot of examples, finding the ODZ allows you to get the answer without bulky layouts, or even verbally.
1. OD3 is an empty set, which means that the original example has no solutions.
1)
2) 3) 
2. B ODZ one or more numbers are found, and a simple substitution quickly determines the roots.
1)
, x=3
2)
Here in the ODZ there is only the number 1, and after substitution it is clear that it is not a root.
3) There are two numbers in the ODZ: 2 and 3, and both are suitable.
4) > In the ODZ there are two numbers 0 and 1, and only 1 is suitable.
ODZ can be used effectively in combination with analysis of the expression itself.
5) 
< ОДЗ: Но в правой части неравенства могут быть только положительные числа, поэтому оставляем х=2. Тогда в неравенство подставим 2.
6) 
From the ODZ it follows that, where we have ..gif" width="143" height="24"> From the ODZ we have: . But then and . Since, there are no solutions.
From the ODZ we have: https://pandia.ru/text/78/083/images/image060_0.gif" width="48" height="24">>, which means . Solving the last inequality, we get x<- 4, что не входит в ОДЗ. Поэтому решения нет.
3) ODZ: . Since then
On the other hand, https://pandia.ru/text/78/083/images/image068_0.gif" width="160" height="24">
ODZ:. Consider the equation on the interval [-1; 0).
It fulfills the following inequalities https://pandia.ru/text/78/083/images/image071_0.gif" width="68" height="24 src=">.gif" width="123" height="24 src="> and there are no solutions. With the function and https://pandia.ru/text/78/083/images/image076_0.gif" width="179" height="25">. ODZ: x>2..gif" width="233" height ="45 src="> Let's find the ODZ:
An integer solution is only possible for x=3 and x=5. By checking we find that the root x=3 does not fit, which means the answer is x=5.
6. Finding the range of acceptable values is extra work. Equivalence of transitions.
You can give examples where the situation is clear even without finding DZ.
1. 
Equality is impossible, because when subtracting a larger expression from a smaller one, the result must be a negative number.
2. 
.
The sum of two non-negative functions cannot be negative.
I will also give examples where finding ODZ is difficult, and sometimes simply impossible.
And finally, searches for ODZ are very often just extra work, which you can do without, thereby proving your understanding of what is happening. There are a huge number of examples that can be given here, so I will choose only the most typical ones. The main solution method in this case is equivalent transformations when moving from one equation (inequality, system) to another.
1.
. ODZ is not needed, because, having found those values of x for which x2 = 1, we cannot obtain x = 0.
2. . ODZ is not needed, because we find out when the radical expression is equal to a positive number.
3. . ODZ is not needed for the same reasons as in the previous example.
4. 
ODZ is not needed, because the radical expression is equal to the square of some function, and therefore cannot be negative.
5. 
6. ..gif" width="271" height="51"> To solve, only one restriction for the radical expression is sufficient. In fact, from the written mixed system it follows that the other radical expression is non-negative.
8. DZ is not needed for the same reasons as in the previous example.
9. 
ODZ is not needed, since it is enough for two of the three expressions under the logarithm signs to be positive to ensure the positivity of the third.
10. 
.gif" width="357" height="51"> ODZ is not needed for the same reasons as in the previous example.
It is worth noting, however, that when solving using the method of equivalent transformations, knowledge of the ODZ (and properties of functions) helps.
Here are some examples.
1. . OD3, which implies that the expression on the right side is positive, and it is possible to write an equation equivalent to this one in this form https://pandia.ru/text/78/083/images/image101_0.gif" width="112" height="27 "> ODZ: But then, and when solving this inequality, it is not necessary to consider the case when the right side is less than 0.
3. . From the ODZ it follows that, and therefore the case when https://pandia.ru/text/78/083/images/image106_0.gif" width="303" height="48"> Go to general view looks like that:
https://pandia.ru/text/78/083/images/image108_0.gif" width="303" height="24">
There are two possible cases: 0
This means that the original inequality is equivalent to the following set of systems of inequalities:
The first system has no solutions, but from the second we obtain: x<-1 – решение неравенства.
Understanding the conditions of equivalence requires knowledge of some subtleties. For example, why are the following equations equivalent:
Or 
And finally, perhaps most importantly. The fact is that equivalence guarantees the correctness of the answer if some transformations of the equation itself are made, but is not used for transformations in only one of the parts. Abbreviations and the use of different formulas in one of the parts are not covered by the equivalence theorems. I have already given some examples of this type. Let's look at some more examples.
1. This decision is natural. On the left side, according to the property of the logarithmic function, we move on to the expression ..gif" width="111" height="48">
Having solved this system, we get the result (-2 and 2), which, however, is not an answer, since the number -2 is not included in the ODZ. So, do we need to install ODS? Of course not. But since we used a certain property of the logarithmic function in the solution, then we are obliged to provide the conditions under which it is satisfied. Such a condition is the positivity of expressions under the logarithm sign..gif" width="65" height="48">.
2. 
..gif" width="143" height="27 src="> numbers are subject to substitution in this way 
. Who wants to do such tedious calculations?.gif" width="12" height="23 src="> add a condition, and you can immediately see that only the number https://pandia.ru/text/78/083/ meets this condition images/image128_0.gif" width="117" height="27 src=">) was demonstrated by 52% of test takers. One of the reasons for such low rates is the fact that many graduates did not select the roots obtained from the equation after squaring it.
3) Consider, for example, the solution to one of the problems C1: “Find all values of x for which the points of the graph of the function 
lie above the corresponding points of the graph of the function ". The task comes down to solving a fractional inequality containing a logarithmic expression. We know the methods for solving such inequalities. The most common of them is the method of intervals. However, when using it, test takers make various mistakes. Let's consider the most common errors using the example of inequality :
X< 10. Они отмечают, что в первом случае решений нет, а во втором – корнями являются числа –1 и . При этом выпускники не учитывают условие x < 10.
8. Conclusion
To summarize, we can say that there is no universal method for solving equations and inequalities. Every time, if you want to understand what you are doing and not act mechanically, a dilemma arises: what solution should you choose, in particular, should you look for ODZ or not? I think that the experience I have gained will help me solve this dilemma. I will stop making mistakes by learning how to use ODZ correctly. Whether I can do this, time, or rather the Unified State Examination, will tell.
9. Literature
And others. “Algebra and the beginnings of analysis 10-11” problem book and textbook, M.: “Prosveshchenie”, 2002. “Handbook of elementary mathematics.” M.: “Nauka”, 1966. Newspaper “Mathematics” No. 46, Newspaper “Mathematics” No. Newspaper “Mathematics” No. “History of mathematics in school grades VII-VIII”. M.: “Prosveshchenie”, 1982. etc. “The most complete edition of versions of real Unified State Examination tasks: 2009/FIPI” - M.: “Astrel”, 2009. etc. “Unified State Examination. Mathematics. Universal materials for preparing students/FIPI" - M.: "Intelligence Center", 2009. etc. "Algebra and the beginnings of analysis 10-11." M.: “Prosveshchenie”, 2007. “Workshop on solving problems in school mathematics (workshop in algebra).” M.: Education, 1976. “25,000 mathematics lessons.” M.: “Enlightenment”, 1993. “Preparing for the Olympiads in mathematics.” M.: “Exam”, 2006. “Encyclopedia for children “MATHEMATICS”” volume 11, M.: Avanta +; 2002. Materials from the sites www. *****, www. *****.
In equations and inequalities of the form , , , , the intersection of the domains of definition of the functions and is called the domain of permissible values (ADV) of the variable, as well as the ARV of the equation or inequality, respectively.
When solving equations (inequalities) with one variable, when the question arises whether to find the ODZ, you can often hear a categorical “yes” and an equally categorical “no”. “First you need to find the ODZ, and then start solving the equation (inequality),” some say. “There is no need to waste time on ODZ; as the solution progresses, we will move on to an equivalent equation (inequality) or to an equivalent system of equations and inequalities or only inequalities. After all, if this is an equation, then a test can be made,” others argue.
So is it possible to find ODZ?
Of course, there is no clear answer to this question. Finding the OD of an equation or inequality is not a mandatory element of the solution. In every specific example this issue is resolved individually.
In some cases, finding the ODZ simplifies the solution of an equation or inequality (examples 1-5), and in some cases it is even a necessary step in the solution (examples 1, 2, 4).
In other cases (examples 6, 7), it is worth abandoning the preliminary finding of the ODZ, since it makes the solution more cumbersome.
Example 1. Solve the equation.
Squaring both sides of the equation will not simplify it, but will complicate it and will not allow us to get rid of radicals. We need to look for another solution.
Let's find the ODZ equation:
Thus, the ODZ contains only one value, and, therefore, only the number 4 can serve as the root of the original equation. By direct substitution we are convinced that is the only root of the equation.
Example 2. Solve the equation.
The presence of radicals of various degrees in the equation - second, third and sixth - makes the solution difficult. Therefore, first of all, let’s find the ODZ equation:
By direct substitution we verify what is the root of the original equation.
Example 3. Solve inequality.
Of course, it is possible to solve this inequality by considering the cases: , , but finding the ODZ immediately simplifies this solution.
ODZ: 
Substituting this single value into the original inequality, we get false numerical inequality. Therefore, the original inequality has no solution.
Answer: no solution.
Example 4. Solve the equation.
Let's write the equation in the form .
An equation of the form is equivalent to a mixed system 
those. 
Of course, finding ODZ here is unnecessary.
In our case we obtain an equivalent system 
those.
The equation is equivalent to the aggregate 
The equation has no rational roots, but it can have irrational roots, finding which will cause difficulties for students. Therefore, we will look for another solution.
Let's return to the original equation and write it in the form .
Let's find the ODZ: .
When the right side of the equation is , and the left side 
. Consequently, the original equation in the range of permissible values of the variable X is equivalent to the system of equations 
the solution of which is only one value.
Thus, in this example, it was the finding of the ODZ that made it possible to solve the original equation.
Example 5. Solve the equation.
Since , and , then when solving the original equation it will be necessary to get rid of modules (open them).
Therefore, first it makes sense to find the ODZ equation:
So, ODZ:
Let's simplify the original equation using the properties of logarithms.
Since in the range of permissible values of the variable X and , then 
, a , then we get an equivalent equation:
Considering that in ODZ , let's move on to the equivalent equation 
and solve it by dividing both sides by 3.
Answer: − 4.75.
Comment.
If one does not find the ODZ, then when solving the equation it would be necessary to consider four cases: , , , . At each of these intervals of constant sign of the expressions under the modulus sign, it would be necessary to expand the moduli and solve the resulting equation. In addition, also perform a check. We see that finding the ODZ of the original equation greatly simplifies its solution.
Example 7. Solve inequality 
.
Since the variable X is also included in the base of the logarithm, then when solving this inequality it will be necessary to consider two cases: and . Therefore, it is impractical to separately find ODZ.
So, let's represent the original inequality in the form 
and it will be equivalent to the combination of two systems:
Answer: 
.
Any expression with a variable has its own range of valid values, where it exists. ODZ must always be taken into account when making decisions. If it is absent, you may get an incorrect result.
This article will show how to correctly find ODZ and use examples. The importance of indicating the DZ when making a decision will also be discussed.
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Valid and invalid variable values
This definition is related to the allowed values of the variable. When we introduce the definition, let's see what result it will lead to.
Starting from grade 7, we begin to work with numbers and numerical expressions. Initial definitions with variables move on to the meaning of expressions with selected variables.
When there are expressions with selected variables, some of them may not satisfy. For example, an expression of the form 1: a, if a = 0, then it does not make sense, since it is impossible to divide by zero. That is, the expression must have values that are suitable in any case and will give an answer. In other words, they make sense with the existing variables.
Definition 1
If there is an expression with variables, then it makes sense only if the value can be calculated by substituting them.
Definition 2
If there is an expression with variables, then it does not make sense when, when substituting them, the value cannot be calculated.
That is, this implies a complete definition
Definition 3
Existing admissible variables are those values for which the expression makes sense. And if it doesn’t make sense, then they are considered unacceptable.
To clarify the above: if there is more than one variable, then there may be a pair of suitable values.
Example 1
For example, consider an expression of the form 1 x - y + z, where there are three variables. Otherwise, you can write it as x = 0, y = 1, z = 2, while another entry has the form (0, 1, 2). These values are called valid, which means that the value of the expression can be found. We get that 1 0 - 1 + 2 = 1 1 = 1. From this we see that (1, 1, 2) are unacceptable. The substitution results in division by zero, that is, 1 1 - 2 + 1 = 1 0.
What is ODZ?
The range of acceptable values is an important element when evaluating algebraic expressions. Therefore, it is worth paying attention to this when making calculations.
Definition 4
ODZ area is the set of values allowed for a given expression.
Let's look at an example expression.
Example 2
If we have an expression of the form 5 z - 3, then the ODZ has the form (− ∞, 3) ∪ (3, + ∞) . This is the range of valid values that satisfies the variable z for a given expression.
If there are expressions of the form z x - y, then it is clear that x ≠ y, z takes any value. This is called ODZ expressions. It must be taken into account so as not to obtain division by zero when substituting.
The range of permissible values and the range of definition have the same meaning. Only the second of them is used for expressions, and the first is used for equations or inequalities. With the help of DL, the expression or inequality makes sense. The domain of definition of the function coincides with the range of permissible values of the variable x for the expression f (x).
How to find ODZ? Examples, solutions
Finding ODZ means finding all valid values suitable for given function or inequality. Failure to meet these conditions may result in incorrect results. To find the ODZ, it is often necessary to go through transformations in a given expression.
There are expressions where their calculation is impossible:
- if there is division by zero;
- taking the root of a negative number;
- the presence of a negative integer indicator – only for positive numbers;
- calculating the logarithm of a negative number;
- domain of definition of tangent π 2 + π · k, k ∈ Z and cotangent π · k, k ∈ Z;
- finding the value of the arcsine and arccosine of a number for a value not belonging to [ - 1 ; 1 ] .
All this shows how important it is to have ODZ.
Example 3
Find the ODZ expression x 3 + 2 x y − 4 .
Solution
Any number can be cubed. This expression does not have a fraction, so the values of x and y can be any. That is, ODZ is any number.
Answer: x and y – any values.
Example 4
Find the ODZ of the expression 1 3 - x + 1 0.
Solution
It can be seen that there is one fraction where the denominator is zero. This means that for any value of x we will get division by zero. This means that we can conclude that this expression is considered undefined, that is, it does not have any additional liability.
Answer: ∅ .
Example 5
Find the ODZ of the given expression x + 2 · y + 3 - 5 · x.
Solution
Availability square root indicates that this expression must be greater than or equal to zero. If it is negative, it has no meaning. This means that it is necessary to write an inequality of the form x + 2 · y + 3 ≥ 0. That is, this is the desired range of acceptable values.
Answer: set of x and y, where x + 2 y + 3 ≥ 0.
Example 6
Determine the ODZ expression of the form 1 x + 1 - 1 + log x + 8 (x 2 + 3) .
Solution
By condition, we have a fraction, so its denominator should not be equal to zero. We get that x + 1 - 1 ≠ 0. The radical expression always makes sense when greater than or equal to zero, that is, x + 1 ≥ 0. Since it has a logarithm, its expression must be strictly positive, that is, x 2 + 3 > 0. The base of the logarithm must also have a positive value and different from 1, then we add the conditions x + 8 > 0 and x + 8 ≠ 1. It follows that the desired ODZ will take the form:
x + 1 - 1 ≠ 0, x + 1 ≥ 0, x 2 + 3 > 0, x + 8 > 0, x + 8 ≠ 1
In other words, it is called a system of inequalities with one variable. The solution will lead to the following ODZ notation [ − 1, 0) ∪ (0, + ∞) .
Answer: [ − 1 , 0) ∪ (0 , + ∞)
Why is it important to consider DPD when driving change?
During identity transformations, it is important to find the ODZ. There are cases when the existence of ODZ does not occur. To understand whether a given expression has a solution, you need to compare the VA of the variables of the original expression and the VA of the resulting one.
Identity transformations:
- may not affect DL;
- may lead to the expansion or addition of DZ;
- can narrow the DZ.
Let's look at an example.
Example 7
If we have an expression of the form x 2 + x + 3 · x, then its ODZ is defined over the entire domain of definition. Even when bringing similar terms and simplifying the expression, the ODZ does not change.
Example 8
If we take the example of the expression x + 3 x − 3 x, then things are different. We have a fractional expression. And we know that division by zero is unacceptable. Then the ODZ has the form (− ∞, 0) ∪ (0, + ∞) . It can be seen that zero is not a solution, so we add it with a parenthesis.
Let's consider an example with the presence of a radical expression.
Example 9
If there is x - 1 · x - 3, then you should pay attention to the ODZ, since it must be written as the inequality (x − 1) · (x − 3) ≥ 0. It is possible to solve by the interval method, then we find that the ODZ will take the form (− ∞, 1 ] ∪ [ 3 , + ∞) . After transforming x - 1 · x - 3 and applying the property of roots, we have that the ODZ can be supplemented and everything can be written in the form of a system of inequalities of the form x - 1 ≥ 0, x - 3 ≥ 0. When solving it, we find that [ 3 , + ∞) . This means that the ODZ is completely written as follows: (− ∞, 1 ] ∪ [ 3 , + ∞) .
Transformations that narrow the DZ must be avoided.
Example 10
Let's consider an example of the expression x - 1 · x - 3, when x = - 1. When substituting, we get that - 1 - 1 · - 1 - 3 = 8 = 2 2 . If we transform this expression and bring it to the form x - 1 · x - 3, then when calculating we find that 2 - 1 · 2 - 3 the expression makes no sense, since the radical expression should not be negative.
It is necessary to adhere to identical transformations that the ODZ will not change.
If there are examples that expand on it, then it should be added to the DL.
Example 11
Let's look at the example of a fraction of the form x x 3 + x. If we cancel by x, then we get that 1 x 2 + 1. Then the ODZ expands and becomes (− ∞ 0) ∪ (0 , + ∞) . Moreover, when calculating, we already work with the second simplified fraction.
In the presence of logarithms, the situation is slightly different.
Example 12
If there is an expression of the form ln x + ln (x + 3), it is replaced by ln (x · (x + 3)), based on the property of the logarithm. From this we can see that the ODZ from (0 , + ∞) to (− ∞ , − 3) ∪ (0 , + ∞) . Therefore, to determine the ODZ ln (x · (x + 3)) it is necessary to carry out calculations on the ODZ, that is, the (0, + ∞) set.
When solving, it is always necessary to pay attention to the structure and type of the expression given by the condition. If the definition area is found correctly, the result will be positive.
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When solving various problems, we very often have to carry out identical transformations of expressions. But it happens that some kind of transformation is acceptable in some cases, but not in others. Significant assistance in terms of monitoring the admissibility of ongoing transformations is provided by ODZ. Let's look at this in more detail.
The essence of the approach is as follows: the ODZ of variables for the original expression is compared with the ODZ of variables for the expression obtained as a result of identical transformations, and based on the comparison results, appropriate conclusions are drawn.
In general, identity transformations can
- do not influence DL;
- lead to the expansion of ODZ;
- lead to a narrowing of ODZ.
Let's illustrate each case with an example.
Consider the expression x 2 +x+3·x, the ODZ of the variable x for this expression is the set R. Now let's do the following identical transformation with this expression - we present similar terms, as a result it will take the form x 2 +4·x. Obviously, the variable x of this expression is also a set R. Thus, the transformation carried out did not change the DZ.
Let's move on. Let's take the expression x+3/x−3/x. In this case, the ODZ is determined by the condition x≠0, which corresponds to the set (−∞, 0)∪(0, +∞) . This expression also contains similar terms, after reducing which we arrive at the expression x, for which the ODZ is R. What we see: as a result of the transformation, the ODZ was expanded (the number zero was added to the ODZ of the variable x for the original expression).
It remains to consider an example of narrowing the range of acceptable values after transformations. Let's take the expression 
. The ODZ of the variable x is determined by the inequality (x−1)·(x−3)≥0, for its solution it is suitable, for example, as a result we have (−∞, 1]∪∪; edited by S. A. Telyakovsky. - 17- ed. - M.: Education, 2008. - 240 pp.: ill. - ISBN 978-5-09-019315-3.
ODZ in logarithmic equations.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
At what point did we fall into the trap of an elementary example? Just at the moment of eliminating logarithms. Logarithms disappeared completely, and along with them the corresponding restrictions on the answer disappeared. Without a trace. In mathematics this is called expansion of ODZ.
So what now, abandon the elimination of logarithms!? Then we won’t be able to solve anything at all... No, we won’t refuse. We'll go the other way! In mathematics this problem is solved this way.
Before solving any logarithmic equation, we write down the ODZ. After that, you can do whatever you want with the equation. I mean, it’s up to you to decide...) Having received the answer, you just need to find out whether the roots are included in the ODZ. Those included are complete, correct solutions. Those that are not included are mercilessly thrown out. These roots were formed during the solution process independently; they are superfluous. They are sometimes called this: extraneous roots.
How to record ODZ?
Very simple. Let's carefully examine the original example. We don’t solve, we don’t transform, exactly inspect, and exactly original! It is important! And it’s easy, too. We look for dangerous places in the example. This division to the expression with an X, even root extraction from the expression with x and logarithms with X's.
We don't know what x is, right? We haven't solved the example yet. But we are firmly convinced that those X's that will give division by zero, taking the square root of a negative number and violating restrictions on logarithms are obviously not suitable as an answer. These X's turn the original example into nonsense. Therefore, such values of x are unacceptable. All other values of x will constitute the ODZ. Range of acceptable values. That's all.
In practice, this is all much easier to do. We read and understand. Let's take the same example:
log 3 (x 2 -3) = log 3 (2x)
Let's look at the example and find out that there are no divisions, there are no roots, but the equation contains expressions with an X inside the logarithm. We recall that the sublogarithmic expression must be always greater than zero. This is how we write it directly:
Note! We Nothing didn't decide! We just wrote it down required condition on All sublogarithmic expression. For everyone logarithm in the example. The system sign (curly brace) indicates that these conditions must be met simultaneously.
That's all. ODZ recorded. Not so difficult, right?
What to do with ODZ?
So, the ODZ was recorded. Half the job is done). What to do next with this recording? This is where we have options.
Option one, universal:
We solve the system of inequalities that we wrote down for the ODZ.
We solve only ODZ! Let’s not touch the example itself for now! We obtain the values of x that are acceptable for this equation. Anyone who knows how to solve systems of inequalities will receive the following answer for our DL:
Those. As an answer, we can only use X's that are larger than the root of three!
That's it, straws have been laid down. Now you can take on the example itself. Feel free to remove logarithms and do any other transformations - we wrote down the original restrictions and saved them.
Having solved the equation itself and received the answers x 1 = 3; x 2 = -1, it is easy to see that only x 1 = 3 is suitable as an answer. The root x 2 = -1 is less than the root of three, it is extraneous. We simply discard it. That's all.
It’s good for those who know how to solve systems of inequalities, right?)
And if with the solution of systems of inequalities, then... not so much? How to be?! How to be, how to be... Learn! But if it really bothers you... Okay, just for you! Method-light.)
Option two, only for simple equations.
So, we have written down the ODZ in the form of a system of inequalities. This system may not be solved. Leave it as is, like this:
And now, one by one We substitute these values into the system of ODZ inequalities.
For x 1 = 3:
We just count and get:
Everything is fine. Both inequalities are true. This means that the troika passes through the ODZ and goes straight back.
Substitute the second root x 2 = -1:
We count and get:
This is categorically false! Minus two is no more than zero! This means that this root is not included in the ODZ. It is simply thrown away and does not lead to any response. All. Note that the root is thrown out if it does not fit at least one system inequality.
Here's the light method. Let me emphasize that this method is simple and obvious. Solving inequalities is replaced by simple calculation. Very good at simple equations. And it is not suitable for logarithmic inequalities. Can you guess why?
Yes, because the answer to inequality usually has not one or two roots, but interval. Those. endless set of numbers. And in the light method you need to substitute in ODZ All meanings... Infinity. Which seems a little difficult, yes...
Here we have looked at just one simple example. But the essence of such work with DZ remains unchanged for any logarithmic equations.
Well, we have dealt with the ODZ - the main trap in logarithmic equations. The most attentive may ask why in the previous lesson we managed without ODZ? Yes, it’s just that the ODZ did not affect the answer in any way! You can check it yourself. It happens. We decided that we didn’t remember about ODZ (or didn’t know at all...), but we still got the correct answer. So - lucky. I’m saying it’s a lottery, if you decide without ODZ...)
And now - attention!
Get into it. And remember one simple thought. This thought will save you from confusion in your decision and confusion in your head:
The solution to any logarithmic equation consists of two equal parts. One part is solving the equation itself. The second is solving the conditions of DL. These parts are resolved regardless from each other. The results are combined at the final stage of the solution.
Keyword Here - "regardless". When solving ODZ, you don’t have to remember about the equation. And vice versa. The main thing is not to forget to compare the results at the very end, throw away the excess, and write down the correct answer.)
Let's sum it up in practical advice.
1. First of all, we write down the conditions of DL according to the original example.
2. We choose where to start the solution. You can start with an equation, or you can start with the conditions of ODZ. We choose what is easier to solve.
3. Having solved the equation and the ODZ, we summarize the results into a general answer.
4. If the example allows, the DL need not be solved. It is enough to substitute the results of the equation into the written conditions of the ODZ and check which solutions pass. Take them for answers.
Well, as usual, we’ll figure it out. There are only a few examples here, but they cover the most popular chips with ODZ. Some tricks (if you see them) allow you to shorten the solution tenfold! I am not kidding.
Find the root or sum of roots (if there are several) of the equations:
log 2 (x 2 +5x-6) = log 2 (4x)
ln(x 3 -7x+2sinx+3) = ln(x 3 -7x+2sinx-4)
Answers (in disarray): 2; there are no solutions; 1; -5.
Well, how is it? I note that it’s scary appearance some examples are deceptive. They can be solved easily.) If you did everything quickly and correctly, you can take on more difficult tasks.
If it didn’t work out, or it took a long time to solve, visit section 555. There these examples are analyzed in detail. The techniques for correct and fast solutions. Sometimes in logarithmic equations half, or even more, do not need to be solved at all. The answer will still be correct. Yes Yes! Section 555 places particular emphasis on this.
Now you can solve simple logarithmic equations quite reliably. Not a lottery, yes...)
And how to reduce complex equations to the simplest, how to use the properties of logarithms and variable substitution to the fullest, how not to fall into an ambush called “Narrowing the ODZ” - all this will be in the following lessons.
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