7th grade

“Identities. Identity transformation of expressions”.

Abdulkerimova Khadizhat Makhmudovna,

mathematic teacher

Lesson Objectives

to acquaint and initially consolidate the concepts of "identically equal expressions", "identity", "identical transformations";

to consider ways to prove identities, to contribute to the development of skills to prove identities;

to check the students' assimilation of the material covered, to form the skills of applying the studied for the perception of the new.

Lesson type: learning new material

Equipment : board, textbook, workbook.

P lan lesson

Organizing time

Checking homework

Knowledge update

The study of new material (Introduction and primary consolidation of the concepts of "identity", "identical transformations").

Training exercises (Formation of the concepts of "identity", "identical transformations").

Reflection of the lesson (Summarize the theoretical information obtained in the lesson).

Homework message (Explain the content of homework)

During the classes

I. Organizational moment.

II . Checking homework. (Front)

III . Knowledge update.

Give an example of a numeric expression and an expression with variables

Compare the values ​​of the expressions x+3 and 3x at x=-4; 1.5; 5

What number cannot be divided by? (0)

Multiplication result? (Work)

Largest two digit number? (99)

What is the product from -200 to 200? (0)

The result of the subtraction. (Difference)

How many grams in a kilogram? (1000)

Commutative property of addition. (The sum does not change from the rearrangement of the places of the terms)

Commutative property of multiplication. (The product does not change from the permutation of the places of factors)

Associative property of addition. (In order to add a number to the sum of two numbers, you can add the sum of the second and third to the first number)

Associative property of multiplication. (to multiply the product of two numbers by the third number, you can multiply the first number by the product of the second and third)

distribution property. (To multiply a number by the sum of two numbers, you can multiply this number by each term and add the results)

IV. Explanation of the new topic:

Find the value of the expressions at x=5 and y=4

3(x+y)=3(5+4)=3*9=27

3x+3y=3*5+3*4=27

We got the same result. It follows from the distributive property that, in general, for any values ​​of the variables, the values ​​of the expressions 3(x + y) and 3x + 3y are equal.

Consider now the expressions 2x + y and 2xy. For x=1 and y=2 they take equal values:

2x+y=2*1+2=4

2xy=2*1*2=4

However, you can specify x and y values ​​such that the values ​​of these expressions are not equal. For example, if x=3, y=4, then

2x+y=2*3+4=10

2xy=2*3*4=24

Definition: Two expressions whose values ​​are equal for any values ​​of the variables are said to be identically equal.

The expressions 3(x+y) and 3x+3y are identically equal, but the expressions 2x+y and 2xy are not identically equal.

The equality 3(x + y) and 3x + 3y is true for any values ​​of x and y. Such equalities are called identities.

Definition: An equality that is true for any values ​​of the variables is called an identity.

True numerical equalities are also considered identities. We have already met with identities. Identities are equalities that express the basic properties of actions on numbers (Students comment on each property by pronouncing it).

a + b = b + a ab=ba (a + b) + c = a + (b + c) (ab)c = a(bc) a(b + c) = ab + ac

Other examples of identities can be given (Students comment on each property, pronouncing it).

a + 0 = a

a * 1 = a

a + (-a) = 0

a * (- b ) = - ab

a - b = a + (- b )

(- a ) * (- b ) = ab

Definition: The replacement of one expression by another, identically equal to it, is called an identical transformation or simply a transformation of an expression.

Teacher:

Identical transformations of expressions with variables are performed based on the properties of operations on numbers.

Identity transformations of expressions are widely used in calculating the values ​​of expressions and solving other problems. You already had to perform some identical transformations, for example, reduction of similar terms, expansion of brackets. Recall the rules for these transformations:

Students:

To bring like terms, it is necessary to add their coefficients and multiply the result by the common letter part;

If there is a plus sign in front of the brackets, then the brackets can be omitted, retaining the sign of each term enclosed in brackets;

If there is a minus sign before the brackets, then the brackets can be omitted by changing the sign of each term enclosed in brackets.

Teacher:

Example 1. We present similar terms

5x + 2x-3x=x(5+2-3)=4x

What rule did we use?

Student:

We have used the rule of reduction of like terms. This transformation is based on the distributive property of multiplication.

Teacher:

Example 2. Expand the brackets in the expression 2a + (b-3 c) = 2 a + b – 3 c

We applied the rule of opening brackets preceded by a plus sign.

Student:

The performed transformation is based on the associative property of addition.

Teacher:

Example 3. Let's open the brackets in the expression a - (4b- c) =a – 4 b + c

We used the rule of opening brackets, which are preceded by a minus sign.

What property is this transformation based on?

Student:

The performed transformation is based on the distributive property of multiplication and the associative property of addition.

V . Doing exercises.

№85 Orally

№86 Orally

№88 Orally

№93

№94

№90av

№96

№97

VI . Lesson reflection .

The teacher asks questions, and the students answer them as they wish.

What two expressions are called identically equal? Give examples.

What equality is called identity? Give an example.

What identical transformations do you know?

VII . Homework . p.5, No. 95, 98,100 (a, c)

Lesson content

Raising a binomial to a power

A binomial is a polynomial that has two terms. In previous lessons, we raised the binomial to the second and third powers, thereby obtaining the abbreviated multiplication formulas:

(a+b) 2 = a 2 + 2ab + b 2

(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3

But the binomial can be raised not only to the second and third powers, but also to the fourth, fifth or higher powers.

For example, let's build a binomial a+b to the fourth degree:

(a+b) 4

We represent this expression as a product of a binomial a+b and the cube of the same binomial

(a+b)(a+b) 3

Factor ( a+b) 3 can be replaced by the right side of the cube formula of the sum of two expressions. Then we get:

(a+b)(a 3 + 3a 2 b + 3ab 2 + b 3)

And this is the usual multiplication of polynomials. Let's execute it:

That is, when constructing a binomial a+b polynomial to the fourth power a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4

(a+b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4

Construction of a binomial a+b to the fourth power, you can also do this: represent the expression ( a+b) 4 as a product of powers (a+b) 2 (a+b) 2

(a+b) 2 (a+b) 2

But the expression ( a+b) 2 is equal to a 2 + 2ab + b 2 . Let's replace in the expression (a+b) 2 (a+b) 2 polynomial sum squares a 2 + 2ab + b 2

(a 2 + 2ab + b 2)(a 2 + 2ab + b 2)

And this is again the usual multiplication of polynomials. Let's execute it. We will get the same result as before:

Raising a trinomial to a power

A trinomial is a polynomial that has three terms. For example, the expression a+b+c is a trinomial.

Sometimes the problem may arise to raise a trinomial to a power. For example, let's square the trinomial a+b+c

(a+b+c) 2

The two terms inside parentheses can be parenthesized. For example, we conclude the sum a+ b in brackets:

((a+b) + c) 2

In this case, the amount a+b will be treated as one member. Then it turns out that we are squaring not a trinomial, but a binomial. Sum a+b will be the first member, and the member c- the second member. And we already know how to square a binomial. To do this, you can use the formula for the square of the sum of two expressions:

(a+b) 2 = a 2 + 2ab + b 2

Let's apply this formula to our example:

In the same way, you can square a polynomial consisting of four or more terms. For example, let's square the polynomial a+b+c+d

(a+b+c+d) 2

We represent the polynomial as the sum of two expressions: a+b and c + d. To do this, enclose them in brackets:

((a+b) + (c + d)) 2

Now we use the formula for the square of the sum of two expressions:

Selection of a full square from a square trinomial

Another identity transformation that can be useful in solving problems is the selection of a full square from a square trinomial.

A square trinomial is a trinomial of the second degree. For example, the following trinomials are square:

The idea of ​​extracting a full square from such trinomials is to represent the original square trinomial as an expression ( a+b) 2 + c, where ( a+b) 2 full square, and c- some numeric or literal expression.

For example, we select the full square from the trinomial 4x 2 + 16x+ 19 .

First you need to build an expression of the form a 2 + 2ab+ b 2 . We will build it from a trinomial 4x 2 + 16x+ 19 . First, let's decide which members will play the role of variables a and b

The role of the variable a dick will play 2 x, since the first term of the trinomial 4x 2 + 16x+ 19 , namely 4 x 2 is obtained if 2 x square:

(2x) 2 = 4x 2

So the variable a equals 2 x

a = 2x

Now we return to the original trinomial and immediately pay attention to the expression 16 x. This expression is twice the product of the first expression a(in our case it is 2 x) and the second yet unknown expression b. Temporarily put a question mark in its place:

2×2 x × ? = 16x

Looking closely at the 2 × 2 expression x × ? = 16x , it becomes intuitively clear that the member b in this situation is the number 4, since the expression 2 × 2 x equals 4 x, and to get 16 x need to multiply 4 x by 4 .

2×2 x × 4 = 16x

From this we conclude that the variable b equals 4

b = 4

So our full square will be the expression (2x) 2 + 2 × 2 x× 4 + 4 2

Now we are all set to extract the full square from the trinomial 4x 2 + 16x+ 19 .

So, back to the original trinomial 4x 2 + 16x+ 19 and try to carefully embed the full square we have obtained into it (2x) 2 + 2 × 2 x× 4 + 4 2

4x 2 + 16x+ 19 =

Instead of 4 x 2 write down (2 x) 2

4x 2 + 16x+ 19 = (2x) 2

4x 2 + 16x+ 19 = (2x) 2 + 2 × 2 x×4

4x 2 + 16x+ 19 = (2x) 2 + 2 × 2 x× 4 + 4 2

And for now, we rewrite member 19 as it is:

4x 2 + 16x + 19 = (2x) 2 + 2 × 2 x× 4 + 4 2 + 19

Now let's pay attention to the fact that the polynomial we have obtained (2x) 2 + 2 × 2 x× 4 + 4 2 + 19 not identical to the original trinomial 4x 2 + 16x+ 19 . You can verify this by bringing the polynomial (2x) 2 + 2 × 2 x× 4 + 4 2 + 19 to standard view:

(2x) 2 + 2 × 2 x× 4 + 4 2 + 19 = 4 x 2 + 16x + 4 2 + 19

We see that we get a polynomial 4x 2 + 16x+ 4 2 + 19 , but it should have turned out 4x 2 + 16x+ 19 . This is due to the fact that term 4 2 was artificially introduced into the original trinomial in order to organize a complete square from the trinomial 4x 2 + 16x+ 19 .

4x 2 + 16x + 19 = (2x) 2 + 2 × 2 x× 4 + 4 2 − 4 2 + 19

Now the expression (2x) 2 + 2 × 2 x× 4 + 4 2 can be collapsed, that is, written in the form ( a+b) 2 . In our case, we get the expression (2 x+ 4) 2

4x 2 + 16x + 19 = (2x) 2 + 2 × 2 x× 4 + 4 2 − 4 2 + 19 = (2x + 4) 2 − 4 2 + 19

The remaining terms −4 2 and 19 can be added. −4 2 is −16 , hence −16 + 19 = 3

4x 2 + 16x + 19 = (2x) 2 + 2 × 2 x× 4 + 4 2 − 4 2 + 19 = (2x + 4) 2 − 4 2 + 19 = (2x+ 4) 2 + 3

Means, 4x 2 + 16x+ 19 = (2x + 4) 2 + 3

Example 2. Select a full square from a square trinomial x 2 + 2x+ 2

First, we construct an expression of the form a 2 + 2 ab+b 2. The role of the variable a in this case x plays because x 2 = x 2 .

The next term of the original trinomial 2 x rewrite in the form of a double product of the first expression (this is our x) and the second expression b(it will be 1).

2× x× 1 = 2 x

If a b= 1 , then the expression will be a perfect square x 2 + 2x+ 1 2 .

Now let's go back to the original square trinomial and embed a full square into it x 2 + 2x+ 1 2

x 2 + 2x+ 2 = x 2 + 2x+ 1 2 − 1 2 + 2 = (x+ 1) 2 + 1

As in the previous example, the member b(in this example, it is 1) was immediately subtracted after the addition in order to preserve the value of the original trinomial.

Consider the following numeric expression:

9 + 6 + 2

The value of this expression is 17

9 + 6 + 2 = 17

Let's try to select a full square in this numerical expression. To do this, we first construct an expression of the form a 2 + 2ab+ b 2 . The role of the variable a in this case, the number 3 plays, since the first term of the expression 9 + 6 + 2, namely 9, can be represented as 3 2 .

We represent the second term 6 as a double product of the first term 3 and the second 1

2 x 3 x 1 = 6

That is a variable b will be equal to one. Then the expression 3 2 + 2 × 3 × 1 + 1 2 will be a perfect square. Let's implement it in the original expression:

− 1 2 + 2

We collapse the full square, and add the terms −1 2 and 2:

3 2 + 6 + 2 = 3 2 + 2 × 3 × 1 + 1 2 − 1 2 + 2 = (3 + 1) 2 + 1

The result is (3 + 1) 2 + 2 , which is still 17

(3 + 1) 2 +1 = 4 2 + 1 = 17

Let's say we have a square and two rectangles. A square with a side of 3 cm, a rectangle with sides of 2 cm and 3 cm, and a rectangle with sides of 1 cm and 2 cm

Calculate the area of ​​each figure. The area of ​​the square will be 3 2 = 9 cm 2, the area of ​​the pink rectangle will be 2 × 3 = 6 cm 2, the area of ​​the lilac one will be 1 × 2 = 2 cm 2

Write the sum of the areas of these rectangles:

9 + 6 + 2

This expression can be understood as the union of a square and two rectangles into a single figure:

Then a figure is obtained, the area of ​​\u200b\u200bwhich is 17 cm 2. Indeed, the presented figure contains 17 squares with a side of 1 cm.

Let's try to form a square from the existing figure. And the largest possible square. To do this, we will use parts from the pink and lilac rectangle.

To form the largest possible square from the existing shape, you can leave the yellow square unchanged, and attach half of the pink rectangle to the bottom of the yellow square:

We see that one more square centimeter is missing before the formation of a complete square. We can take it from the lilac rectangle. So, let's take one square from the lilac rectangle and attach it to the formed large square:

Now let's take a closer look at what we have come to. Namely, on the yellow part of the figure and the pink part, which essentially increased the previous yellow square. Doesn't this mean that there was a side of the square equal to 3 cm, and this side was increased by 1 cm, which ultimately led to an increase in area?

(3 + 1) 2

The expression (3 + 1) 2 is 16 because 3 + 1 = 4 and 4 2 = 16 . The same result can be obtained using the formula for the square of the sum of two expressions:

(3 + 1) 2 = 3 2 + 6 + 1 = 9 + 6 + 1 = 16

Indeed, the resulting square contains 16 squares.

The remaining one square from the lilac rectangle can be attached to the resulting large square. After all, it was originally about a single figure:

(3 + 1) 2 + 1

Attaching a small square to an existing large square is described by the expression (3 + 1) 2 + 1 . And this is the selection of the full square from the expression 9 + 6 + 2

9 + 6 + 2 = 3 2 + 6 + 2 = 3 2 + 2 × 3 × 1 + 1 2 − 1 2 + 2 = (3 + 1) 2 + 1

The expression (3 + 1) 2 + 1 , like the expression 9 + 6 + 2 , is equal to 17 . Indeed, the area of ​​the resulting figure is 17 cm 2.

Example 4. Let's perform the selection of the full square from the square trinomial x 2 + 6x + 8

x 2 + 6x + 8 = x 2+2× x× 3 + 3 2 − 3 2 + 8 = ( x + 3) 2 − 1

In some examples, when building an expression a 2 + 2ab+ b 2 it is not possible to immediately determine the values ​​of variables a and b .

For example, let's perform the extraction of a full square from a square trinomial x 2 + 3x+ 2

variable a corresponds x. Second Member 3 x cannot be represented as a double product of the first expression and the second. In this case, the second term should be multiplied by 2, and so that the value of the original polynomial does not change, immediately divide by 2. It will look like this.

Let two algebraic expressions be given:

Let's make a table of the values ​​of each of these expressions for different numerical values ​​of the letter x.

We see that for all those values ​​that were given to the letter x, the values ​​of both expressions turned out to be equal. The same will be true for any other value of x.

To verify this, we transform the first expression. Based on the distribution law, we write:

Having performed the indicated operations on the numbers, we get:

So, the first expression, after its simplification, turned out to be exactly the same as the second expression.

Now it is clear that for any value of x, the values ​​of both expressions are equal.

Expressions whose values ​​are equal for any values ​​of the letters included in them are called identically equal or identical.

Hence, they are identical expressions.

Let's make one important remark. Let's take expressions:

Having compiled a table similar to the previous one, we will make sure that both expressions, for any value of x, except for have equal numerical values. Only when the second expression is equal to 6, and the first loses its meaning, since the denominator is zero. (Recall that you cannot divide by zero.) Can we say that these expressions are identical?

We agreed earlier that each expression will be considered only for admissible values ​​of letters, that is, for those values ​​for which the expression does not lose its meaning. This means that here, when comparing two expressions, we take into account only those letter values ​​that are valid for both expressions. Therefore, we must exclude the value. And since for all other values ​​of x both expressions have the same numerical value, we have the right to consider them identical.

Based on what has been said, we give the following definition of identical expressions:

1. Expressions are called identical if they have the same numerical values ​​for all admissible values ​​of the letters included in them.

If we connect two identical expressions with an equal sign, then we get an identity. Means:

2. An identity is an equality that is true for all admissible values ​​of the letters included in it.

We have already encountered identities before. So, for example, all equalities are identities, with which we expressed the basic laws of addition and multiplication.

For example, equalities expressing the commutative law of addition

and the associative law of multiplication

are valid for any values ​​of letters. Hence, these equalities are identities.

All true arithmetic equalities are also considered identities, for example:

In algebra, one often has to replace an expression with another that is identical to it. Let, for example, it is required to find the value of the expression

We will greatly facilitate the calculations if we replace the given expression with an expression that is identical to it. Based on the distribution law, we can write:

But the numbers in brackets add up to 100. So, we have an identity:

Substituting 6.53 instead of a on the right side of it, we immediately (in the mind) find the numerical value (653) of this expression.

Replacing one expression with another, identical to it, is called the identical transformation of this expression.

Recall that any algebraic expression for any admissible values ​​of letters is some

number. It follows from this that all the laws and properties of arithmetic operations that were given in the previous chapter are applicable to algebraic expressions. So, the application of the laws and properties of arithmetic operations transforms a given algebraic expression into an expression that is identical to it.

In the course of studying algebra, we came across the concepts of polynomial (for example ($y-x$ ,$\ 2x^2-2x$ and so on) and algebraic fraction (for example $\frac(x+5)(x)$ , $\frac(2x ^2)(2x^2-2x)$,$\ \frac(x-y)(y-x)$ etc.) The similarity of these concepts is that both in polynomials and in algebraic fractions there are variables and numerical values, arithmetic actions: addition, subtraction, multiplication, exponentiation The difference between these concepts is that in polynomials division by a variable is not performed, and in algebraic fractions division by a variable can be performed.

Both polynomials and algebraic fractions are called rational algebraic expressions in mathematics. But polynomials are integer rational expressions, and algebraic fractional expressions are fractionally rational expressions.

It is possible to obtain a whole algebraic expression from a fractional-rational expression using the identical transformation, which in this case will be the main property of a fraction - reduction of fractions. Let's check it out in practice:

Example 1

Transform:$\ \frac(x^2-4x+4)(x-2)$

Solution: This fractional-rational equation can be transformed by using the basic property of the fraction-cancellation, i.e. dividing the numerator and denominator by the same number or expression other than $0$.

This fraction cannot be reduced immediately, it is necessary to convert the numerator.

We transform the expression in the numerator of the fraction, for this we use the formula for the square of the difference: $a^2-2ab+b^2=((a-b))^2$

The fraction has the form

\[\frac(x^2-4x+4)(x-2)=\frac(x^2-4x+4)(x-2)=\frac(((x-2))^2)( x-2)=\frac(\left(x-2\right)(x-2))(x-2)\]

Now we see that there is a common factor in the numerator and denominator - this is the expression $x-2$, on which we will reduce the fraction

\[\frac(x^2-4x+4)(x-2)=\frac(x^2-4x+4)(x-2)=\frac(((x-2))^2)( x-2)=\frac(\left(x-2\right)(x-2))(x-2)=x-2\]

After reduction, we have obtained that the original fractional-rational expression $\frac(x^2-4x+4)(x-2)$ has become a polynomial $x-2$, i.e. whole rational.

Now let's pay attention to the fact that the expressions $\frac(x^2-4x+4)(x-2)$ and $x-2\ $ can be considered identical not for all values ​​of the variable, because in order for a fractional-rational expression to exist and for the reduction by the polynomial $x-2$ to be possible, the denominator of the fraction should not be equal to $0$ (as well as the factor by which we reduce. In this example, the denominator and factor are the same, but this is not always the case).

Variable values ​​for which the algebraic fraction will exist are called valid variable values.

We put a condition on the denominator of the fraction: $x-2≠0$, then $x≠2$.

So the expressions $\frac(x^2-4x+4)(x-2)$ and $x-2$ are identical for all values ​​of the variable except $2$.

Definition 1

identically equal Expressions are those that are equal for all possible values ​​of the variable.

An identical transformation is any replacement of the original expression with an identically equal one. Such transformations include performing actions: addition, subtraction, multiplication, taking a common factor out of the bracket, bringing algebraic fractions to a common denominator, reducing algebraic fractions, bringing like terms, etc. It must be taken into account that a number of transformations, such as reduction, reduction of similar terms, can change the allowable values ​​of the variable.

Techniques used to prove identities

Convert the left side of the identity to the right side or vice versa using identity transformations

Reduce both parts to the same expression using identical transformations

Transfer the expressions in one part of the expression to another and prove that the resulting difference is equal to $0$

Which of the above methods to use to prove a given identity depends on the original identity.

Example 2

Prove the identity $((a+b+c))^2- 2(ab+ac+bc)=a^2+b^2+c^2$

Solution: To prove this identity, we use the first of the above methods, namely, we will transform the left side of the identity until it is equal to the right side.

Consider the left side of the identity: $\ ((a+b+c))^2- 2(ab+ac+bc)$- it is the difference of two polynomials. In this case, the first polynomial is the square of the sum of three terms. To square the sum of several terms, we use the formula:

\[((a+b+c))^2=a^2+b^2+c^2+2ab+2ac+2bc\]

To do this, we need to multiply a number by a polynomial. Recall that for this we need to multiply the common factor outside the brackets by each term of the polynomial in brackets. Then we get:

$2(ab+ac+bc)=2ab+2ac+2bc$

Now back to the original polynomial, it will take the form:

$((a+b+c))^2- 2(ab+ac+bc)=\ a^2+b^2+c^2+2ab+2ac+2bc-(2ab+2ac+2bc)$

Note that there is a “-” sign in front of the bracket, which means that when the brackets are opened, all the signs that were in the brackets are reversed.

$((a+b+c))^2- 2(ab+ac+bc)=\ a^2+b^2+c^2+2ab+2ac+2bc-(2ab+2ac+2bc)= a ^2+b^2+c^2+2ab+2ac+2bc-2ab-2ac-2bc$

If we bring similar terms, then we get that the monomials $2ab$, $2ac$,$\ 2bc$ and $-2ab$,$-2ac$, $-2bc$ cancel each other out, i.e. their sum is equal to $0$.

$((a+b+c))^2- 2(ab+ac+bc)=\ a^2+b^2+c^2+2ab+2ac+2bc-(2ab+2ac+2bc)= a ^2+b^2+c^2+2ab+2ac+2bc-2ab-2ac-2bc=a^2+b^2+c^2$

So, by identical transformations, we obtained the identical expression on the left side of the original identity

$((a+b+c))^2- 2(ab+ac+bc)=\ a^2+b^2+c^2$

Note that the resulting expression shows that the original identity is true.

Note that in the original identity, all values ​​of the variable are allowed, which means that we have proved the identity using identical transformations, and it is true for all allowed values ​​of the variable.

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Slides captions:

Identities. Identity transformations of expressions. 7th grade.

Find the value of the expressions at x=5 and y=4 3(x+y)= 3(5+4)=3*9=27 3x+3y= 3*5+3*4=27 Find the value of the expressions at x=6 and y=5 3(x+y)= 3(6+5)=3*11=33 3x+3y= 3*6+3*5=33

CONCLUSION: We got the same result. It follows from the distributive property that, in general, for any values ​​of the variables, the values ​​of the expressions 3(x + y) and 3x + 3y are equal. 3(x+y) = 3x+3y

Consider now the expressions 2x + y and 2xy. for x=1 and y=2 they take equal values: 2x+y=2*1+2=4 2x=2*1*2=4 for x=3, y=4 expression values ​​are different 2x+y=2* 3+4=10 2xy=2*3*4=24

CONCLUSION: The expressions 3(x+y) and 3x+3y are identically equal, but the expressions 2x+y and 2xy are not identically equal. Definition: Two expressions whose values ​​are equal for any values ​​of the variables are said to be identically equal.

IDENTITY The equality 3(x+y) and 3x+3y is true for any values ​​of x and y. Such equalities are called identities. Definition: An equality that is true for any values ​​of the variables is called an identity. True numerical equalities are also considered identities. We have already met with identities.

Identities are equalities expressing the basic properties of actions on numbers. a + b = b + a ab = ba (a + b) + c = a + (b + c) (ab)c = a(bc) a(b + c) = ab + ac

Other examples of identities can be given: a + 0 = a a * 1 = a a + (-a) = 0 a * (- b) = - ab a- b = a + (- b) (-a) * ( -b) = ab Replacing one expression with another expression identically equal to it is called identity transformation or simply expression transformation.

To bring like terms, you need to add their coefficients and multiply the result by the common letter part. Example 1. We give like terms 5x + 2x-3x \u003d x (5 + 2-3) \u003d 4x

If there is a plus sign in front of the brackets, then the brackets can be omitted, preserving the sign of each term enclosed in brackets. Example 2. Expand the brackets in the expression 2a + (b -3 c) = 2 a + b - 3 c

If there is a minus sign before the brackets, then the brackets can be omitted by changing the sign of each term enclosed in brackets. Example 3. Let's open the brackets in the expression a - (4 b - c) \u003d a - 4 b + c

Homework: p. 5, No. 91, 97, 99 Thank you for the lesson!

On the topic: methodological developments, presentations and notes

Methods of preparing students for the exam in the section "Expressions and transformation of expressions"

This project was developed with the aim of preparing students for state exams in grade 9 and later on for a unified state exam in grade 11....