How to find valid values ​​for a variable in an expression. Acceptable range - ODZ

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48. Types of algebraic expressions.

From numbers and variables, using the signs of addition, subtraction, multiplication, division, raising to a rational power and extracting the root, and using brackets, algebraic expressions are compiled.

Examples of algebraic expressions:

If an algebraic expression does not contain division into variables and extraction of a root from variables (in particular, exponentiation with a fractional exponent), then it is called an integer. Of the ones written above, expressions 1, 2 and 6 are integers.

If an algebraic expression is composed of numbers and variables using the operations of addition, subtraction, multiplication, exponentiation with a natural exponent and division, and division into expressions with variables is used, then it is called fractional. So, from the expressions written above, expressions 3 and 4 are fractional.

Integer and fractional expressions are called rational expressions. So, from the above, rational expressions are expressions 1, 2, 3, 4 and 6.

If an algebraic expression uses the extraction of a root from variables (or the raising of variables to a fractional power), then such an algebraic expression is called irrational. So, from the above, expressions 5 and 7 are irrational.

So, algebraic expressions can be rational and irrational. Rational expressions, in turn, are divided into integer and fractional.

49. Valid values ​​of variables. Domain of definition of an algebraic expression.

Variable values ​​for which the algebraic expression makes sense are called admissible variable values. The set of all admissible values ​​of variables is called the domain of the algebraic expression.

An integer expression makes sense for any values ​​of its variables. So, for any values ​​of the variables, the whole expressions 1, 2, 6 from paragraph 48 make sense.

Fractional expressions do not make sense for those values ​​​​of variables that turn the denominator to zero. So, the fractional expression 3 from item 48 makes sense for all o, except for , and the fractional expression 4 makes sense for all a, b, c, except for the values ​​a

An irrational expression does not make sense for those values ​​of variables that turn into a negative number an expression contained under the root sign of an even degree or under the sign of exponentiation to a fractional power. Thus, the irrational expression 5 makes sense only for those a, b for which a the irrational expression 7 makes sense only for and (see item 48).

If in an algebraic expression the variables are given valid values, then a numerical expression will be obtained; its value is called the value of the algebraic expression for the chosen values ​​of the variables.

Example. Find the value of the expression when

Decision. We have

50. The concept of the identical transformation of an expression. Identity.

Consider two expressions When we have . The numbers 0 and 3 are called the corresponding values. expressions for Let's find the corresponding values ​​of the same expressions for

The corresponding values ​​of two expressions can be equal to each other (for example, in the considered example, the equality is satisfied), or they can differ from each other (for example, in the considered example).

Any expression with a variable has its range of valid values, where it exists. DHS must always be taken into account in the decision. If not, you may get an incorrect result.

This article will show how to correctly find the ODZ, use it with examples. It will also consider the importance of specifying the ODZ in the decision.

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Valid and invalid variable values

This definition is related to the allowed values ​​of the variable. When introducing a 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 jump to the value of expressions with selected variables.

When there are expressions with selected variables, some of them may not satisfy. For example, an expression like 1: a, if a \u003d 0, then it does not make sense, since it is impossible to divide by zero. That is, the expression should have such values ​​that will fit in any case and give the answer. In other words, they make sense with the available variables.

Definition 1

If there is an expression with variables, then it makes sense only if, when they are substituted, the value can be calculated.

Definition 2

If there is an expression with variables, then it does not make sense when, with their substitution, the value cannot be calculated.

That is, from this follows the complete definition

Definition 3

Existing valid variables are those values ​​for which the expression makes sense. And if it makes no sense, then they are considered invalid.

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 like 1 x - y + z , where there are three variables. Otherwise, you can write it as x = 0 , y = 1 , z = 2 , while the other notation is (0 , 1 , 2) . These values ​​are called valid, which means that you can find the value of the expression. We get that 1 0 - 1 + 2 = 1 1 = 1 . From here we see that (1 , 1 , 2) are invalid. The substitution results in division by zero, that is, 1 1 - 2 + 1 = 1 0 .

What is ODZ?

Valid range - important element when evaluating algebraic expressions. Therefore, it is worth paying attention to this when calculating.

Definition 4

ODZ area is the set of values ​​allowed for the given expression.

Let's take an example of an 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 the 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 what is called the ODZ expression. It must be taken into account in order not to get a division by zero when substituting.

The range of valid values ​​and the domain of definition have the same meaning. Only the second of them is used for expressions, and the first one is used for equations or inequalities. With the help of DPV, the expression or inequality makes sense. The domain of the function definition coincides with the domain of admissible values ​​of the variable x to the expression f (x) .

How to find ODZ? Examples, Solutions

To find the DPV means to find all valid values ​​that fit a given function or inequality. If these conditions are not met, an incorrect result can be obtained. To find the ODZ, it is often necessary to go through transformations in a given expression.

There are expressions where they cannot be evaluated:

• if there is a division by zero;
• extracting the root of a negative number;
• the presence of a negative integer indicator - only for positive numbers;
• calculating the logarithm of a negative number;
• the domain of definition of the tangent π 2 + π · k , k ∈ Z and the cotangent π · k , k ∈ Z ;
• finding the value of the arcsine and arccosine of a number with a value that does not belong to [ - 1 ; one ] .

All of this speaks to the importance of having a DHS.

Example 3

Find the ODZ expression x 3 + 2 x y − 4 .

Decision

Any number can be cubed. This expression does not have a fraction, so x and y can be anything. That is, ODZ is any number.

Answer: x and y are any values.

Example 4

Find the ODZ expression 1 3 - x + 1 0 .

Decision

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 a division by zero. This means that we can conclude that this expression is considered to be indefinite, that is, it does not have ODZ.

Answer: ∅ .

Example 5

Find the ODZ of the given expression x + 2 · y + 3 - 5 · x .

Decision

The presence of a square root indicates that this expression must be greater than or equal to zero. If it is negative, it has no meaning. Hence, it is necessary to write down 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) .

Decision

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, i.e. 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 positive value and different from 1 , then we add more conditions x + 8 > 0 and x + 8 ≠ 1 . From this 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 such a record of the ODZ [ − 1 , 0) ∪ (0 , + ∞) .

Answer: [ − 1 , 0) ∪ (0 , + ∞)

Why is it important to take LHS into account when making changes?

For identical transformations, it is important to find the ODZ. There are cases when the existence of ODZ does not take place. To understand whether the solution has a given expression, you need to compare the ODZ of the variables of the original expression and the ODZ of the received expression.

Identity transformations:

• may not affect ODZ;
• may lead to an extension or addition to the DHS;
• can narrow the ODZ.

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 on the entire domain of definition. Even with the reduction of similar terms and simplification of 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 not allowed. 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.

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 an inequality (x − 1) · (x − 3) ≥ 0 . It is possible to solve by the interval method, then we get that the ODZ will take the form (− ∞ , 1 ] ∪ [ 3 , + ∞) . After transforming x - 1 · x - 3 and applying the properties of the roots, we have that the ODZ can be supplemented and written down as a system of inequalities of the form x - 1 ≥ 0 , x - 3 ≥ 0 . When solving it, we obtain that [ 3 , + ∞) . Hence, the ODZ is written in full as follows: (− ∞ , 1 ] ∪ [ 3 , + ∞) .

Changes that narrow the DHS should be avoided.

Example 10

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 this expression is transformed and brought to the form x - 1 x - 3, then when calculating we get that 2 - 1 2 - 3 the expression does not make sense, since the radical expression should not be negative.

Identical transformations should be followed, which will not change the DHS.

If there are examples that extend it, then it should be added to the DPV.

Example 11

Consider the example of a fraction of the form x x 3 + x. If we reduce by x , then we get that 1 x 2 + 1 . Then the ODZ expands and becomes (− ∞ 0) ∪ (0 , + ∞) . Moreover, when calculating, we are already working 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. This shows that the ODZ from (0 , + ∞) to (− ∞ , − 3) ∪ (0 , + ∞) . Therefore, to determine the ODZ ln (x (x + 3)) it is necessary to perform calculations on the ODZ, that is, (0 , + ∞) sets.

When solving, it is always necessary to pay attention to the structure and form of the expression given by the condition. If the domain of definition is found correctly, the result will be positive.

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This lesson discusses the concept of an algebraic fraction. With fractions, a person meets in the simplest life situations: when it is necessary to divide an object into several parts, for example, to cut a cake equally for ten people. Obviously, everyone will get a piece of the cake. In this case, we are faced with the concept of a numerical fraction, but a situation is possible when an object is divided into an unknown number of parts, for example, by x. In this case, the concept of a fractional expression arises. You already met with integer expressions (not containing division into expressions with variables) and their properties in grade 7. Next, we will consider the concept of a rational fraction, as well as the allowable values ​​of variables.

Rational expressions are divided into integer and fractional expressions.

Definition.rational fraction is a fractional expression of the form , where are polynomials. - numerator denominator.

Examplesrational expressions:- fractional expressions; are integer expressions. In the first expression, for example, the numerator is , and the denominator is .

Meaning algebraic fraction, like any algebraic expression, depends on numerical value the variables it contains. In particular, in the first example the value of the fraction depends on the values ​​of the variables and , and in the second only on the value of the variable .

Consider the first typical task: calculating the value rational fraction at different values variables included in it.

Example 1 Calculate the value of the fraction for a), b), c)

Decision. Substitute the values ​​of the variables into the indicated fraction: a), b), c) - does not exist (because you cannot divide by zero).

Answer: a) 3; b) 1; c) does not exist.

As you can see, there are two typical problems for any fraction: 1) calculating the fraction, 2) finding valid and invalid values literal variables.

Definition.Valid Variable Values are the values ​​of the variables for which the expression makes sense. The set of all admissible values ​​of variables is called ODZ or domain.

The value of literal variables may be invalid if the denominator of the fraction for these values zero. In all other cases, the values ​​of the variables are valid, since the fraction can be calculated.

Example 2

Decision. For this expression to make sense, it is necessary and sufficient that the denominator of the fraction does not equal zero. Thus, only those values ​​of the variable for which the denominator will be equal to zero will be invalid. The denominator of the fraction, so we solve the linear equation:

Therefore, for the value of the variable, the fraction does not make sense.

Answer: -5.

From the solution of the example, the rule for finding invalid values ​​of variables follows - the denominator of the fraction is equal to zero and the roots of the corresponding equation are found.

Let's look at a few similar examples.

Example 3 Determine at what values ​​of a variable a fraction does not make sense .

Decision..

Answer..

Example 4 Determine for what values ​​of the variable the fraction does not make sense.

Decision..

There are other formulations of this problem - to find domain or range of valid expression values ​​(ODZ). This means - find all valid values ​​of variables. In our example, these are all values ​​except . The domain of definition is conveniently depicted on the numerical axis.

To do this, we will cut out a point on it, as shown in the figure:

Rice. one

Thus, fraction domain will be all numbers except 3.

Answer..

Example 5 Determine for what values ​​of the variable the fraction does not make sense.

Decision..

Let's depict the resulting solution on the numerical axis:

Rice. 2

Answer..

Example 6

Decision.. We have obtained the equality of two variables, we give numerical examples: or etc.

Let's plot this solution on a graph in the Cartesian coordinate system:

Rice. 3. Graph of a function

The coordinates of any point lying on this graph are not included in the area of ​​​​admissible values ​​of the fraction.

Answer..

In the considered examples, we were faced with a situation where a division by zero occurred. Now consider the case when there is more interesting situation with division type .

Example 7 Determine for what values ​​of the variables the fraction does not make sense.

Decision..

It turns out that the fraction does not make sense when . But it can be argued that this is not the case, because: .

It may seem that if the final expression is equal to 8 for , then the original expression can also be calculated, and, therefore, makes sense for . However, if we substitute it into the original expression, we get - it does not make sense.

Answer..

To understand this example in more detail, we solve the following problem: for what values ​​is the indicated fraction equal to zero?