What is called the identity transformation of the expression. Converting expressions

The arithmetic action that is performed last when calculating the value of the expression is the "main" one.

That is, if you substitute any (any) numbers instead of letters, and try to calculate the value of the expression, then if the last action is multiplication, then we have a product (the expression is factorized).

If the last action is addition or subtraction, this means that the expression is not factorized (and therefore cannot be canceled).

To fix the solution yourself, take a few examples:

Examples:

Solutions:

1. I hope you didn’t rush to cut and? It was still not enough to "cut" units like this:

The first step is to factorize:

4. Addition and subtraction of fractions. Bringing fractions to a common denominator.

Adding and subtracting ordinary fractions is a very familiar operation: we look for a common denominator, multiply each fraction by the missing factor and add / subtract the numerators.

Let's remember:

Answers:

1. The denominators and are mutually prime, that is, they have no common factors. Therefore, the LCM of these numbers is equal to their product. This will be the common denominator:

2. Here the common denominator is:

3. Here, first of all, we turn the mixed fractions into incorrect ones, and then - according to the usual scheme:

It is quite another matter if the fractions contain letters, for example:

Let's start simple:

a) Denominators do not contain letters

Here everything is the same as with ordinary numeric fractions: find the common denominator, multiply each fraction by the missing factor and add / subtract the numerators:

now in the numerator you can bring similar ones, if any, and decompose into factors:

Try it yourself:

Answers:

b) Denominators contain letters

Let's remember the principle of finding a common denominator without letters:

· First of all, we determine the common factors;

· Then write out all common factors one time;

· And multiply them by all other factors that are not common.

To determine the common factors of the denominators, we first decompose them into prime factors:

Let's emphasize the common factors:

Now let's write out the common factors one time and add to them all non-common (not underlined) factors:

This is the common denominator.

Let's go back to the letters. The denominators are shown in exactly the same way:

· We decompose the denominators into factors;

· We determine common (identical) factors;

· Write out all common factors once;

· We multiply them by all other factors, not common.

So, in order:

1) we decompose the denominators into factors:

2) we determine the common (identical) factors:

3) we write out all the common factors one time and multiply them by all the other (unstressed) factors:

So the common denominator is here. The first fraction must be multiplied by, the second by:

By the way, there is one trick:

For instance: .

We see the same factors in the denominators, only all with different indicators. The common denominator will be:

to the extent

to the extent

to the extent

in degree.

Let's complicate the task:

How do you make fractions the same denominator?

Let's remember the basic property of a fraction:

Nowhere is it said that the same number can be subtracted (or added) from the numerator and denominator of a fraction. Because this is not true!

See for yourself: take any fraction, for example, and add some number to the numerator and denominator, for example. What has been learned?

So, another unshakable rule:

When bringing fractions to a common denominator, use only multiplication!

But what must be multiplied by in order to receive?

Here on and multiply. And multiply by:

Expressions that cannot be factorized will be called “elementary factors”.

For example, is an elementary factor. - too. But - no: it is factorized.

What do you think about expression? Is it elementary?

No, since it can be factorized:

(you already read about factorization in the topic "").

So, the elementary factors into which you expand the expression with letters are analogous to the prime factors into which you expand the numbers. And we will deal with them in the same way.

We see that there is a factor in both denominators. It will go to the common denominator in power (remember why?).

The factor is elementary, and it is not common for them, which means that the first fraction will simply have to be multiplied by it:

Another example:

Solution:

Before multiplying these denominators in a panic, you need to think about how to factor them? They both represent:

Fine! Then:

Another example:

Solution:

As usual, factor the denominators. In the first denominator, we simply put it outside the brackets; in the second - the difference of squares:

It would seem that there are no common factors. But if you look closely, then they are so similar ... And the truth:

So we will write:

That is, it turned out like this: inside the parenthesis, we swapped the terms, and at the same time the sign in front of the fraction changed to the opposite. Take note, you will have to do this often.

Now we bring to a common denominator:

Got it? Let's check it out now.

Tasks for independent solution:

Answers:

Here we must remember one more - the difference between the cubes:

Note that the denominator of the second fraction is not the "square of the sum" formula! The square of the sum would look like this:.

A is the so-called incomplete square of the sum: the second term in it is the product of the first and the last, and not their doubled product. The incomplete square of the sum is one of the factors in the decomposition of the difference of cubes:

What if there are already three fractions?

The same thing! First of all, we will do so that the maximum number of factors in the denominators is the same:

Pay attention: if you change the signs within one parenthesis, the sign in front of the fraction changes to the opposite. When we change the signs in the second parenthesis, the sign in front of the fraction is reversed again. As a result, it (the sign in front of the fraction) has not changed.

In the common denominator, write out the first denominator in full, and then add to it all the factors that have not yet been written, from the second, and then from the third (and so on, if there are more fractions). That is, it turns out like this:

Hmm ... With fractions, it’s clear what to do. But what about the deuce?

It's simple: you can add fractions, right? This means that we need to make the deuce become a fraction! Remember: a fraction is a division operation (the numerator is divided by the denominator, in case you suddenly forgot). And there is nothing easier than dividing a number by. In this case, the number itself will not change, but it will turn into a fraction:

Exactly what is needed!

5. Multiplication and division of fractions.

Well, the hardest part is over now. And ahead of us is the simplest, but at the same time the most important:

Procedure

What is the procedure for calculating a numeric expression? Remember by counting the meaning of such an expression:

Did you count it?

It should work.

So, let me remind you.

The first step is to calculate the degree.

The second is multiplication and division. If there are several multiplications and divisions at the same time, you can do them in any order.

And finally, we do addition and subtraction. Again, in any order.

But: the expression in parentheses is evaluated out of order!

If several brackets are multiplied or divided by each other, we first calculate the expression in each of the brackets, and then we multiply or divide them.

What if there are more brackets inside the brackets? Well, let's think about it: some expression is written inside the brackets. And when evaluating an expression, what is the first thing to do? That's right, calculate the parentheses. Well, we figured it out: first we calculate the inner brackets, then everything else.

So, the procedure for the expression above is as follows (the current action is highlighted in red, that is, the action that I am performing right now):

Okay, it's all simple.

But this is not the same as an expression with letters?

No, it's the same! Only instead of arithmetic operations, you need to do algebraic ones, that is, the actions described in the previous section: bringing similar, addition of fractions, reduction of fractions, and so on. The only difference is the effect of factoring polynomials (we often use it when working with fractions). Most often, for factoring, you need to use i or just put the common factor outside the parentheses.

Usually our goal is to present an expression in the form of a work or a particular.

For instance:

Let's simplify the expression.

1) The first is to simplify the expression in parentheses. There we have the difference of fractions, and our goal is to present it as a product or quotient. So, we bring the fractions to a common denominator and add:

It is impossible to simplify this expression anymore, all the factors here are elementary (do you still remember what this means?).

2) We get:

Multiplication of fractions: what could be easier.

3) Now you can shorten:

OK it's all over Now. Nothing complicated, right?

Another example:

Simplify the expression.

First try to solve it yourself, and only then see the solution.

Solution:

First of all, let's define the order of actions.

First, we add the fractions in brackets, we get one instead of two fractions.

Then we will divide the fractions. Well, add the result with the last fraction.

I will schematically number the steps:

Now I will show the whole process, coloring the current action in red:

1. If there are similar ones, they must be brought immediately. At whatever moment we have similar ones, it is advisable to bring them right away.

2. The same applies to the reduction of fractions: as soon as there is an opportunity to reduce, it must be used. The exception is fractions that you add or subtract: if they now have the same denominators, then the reduction should be left for later.

Here are some tasks for you to solve on your own:

And promised at the very beginning:

Answers:

Solutions (concise):

If you have coped with at least the first three examples, then you have mastered the topic.

Now forward to learning!

TRANSFORMATION OF EXPRESSIONS. SUMMARY AND BASIC FORMULAS

Basic simplification operations:

• Bringing similar: to add (bring) such terms, you need to add their coefficients and assign the letter part.
• Factorization: factoring out the common factor, application, etc.
• Fraction reduction: the numerator and denominator of a fraction can be multiplied or divided by the same non-zero number, which does not change the value of the fraction.
  1) numerator and denominator factor
  2) if there are common factors in the numerator and denominator, they can be crossed out.

  IMPORTANT: only multipliers can be reduced!

• Addition and subtraction of fractions:
  ;
• Multiplication and division of fractions:
  ;

Identical transformations

1. The concept of identity. The main types of identical transformations and the stages of their study.

11 learning various transformations of expressions and formulas takes up a poor part of the study time in the school mathematics course. The simplest ^ "" education, based on the properties of arithmetic operations, is already in elementary school. But the main load on the formation of skills and abilities to perform transformations is borne by the course of school algebra 1> then it is connected:

with a sharp increase in the number of transformations being made, their variability;

with the complication of activities to justify them and clarify the conditions of applicability;

i) with the isolation and study of the generalized concepts of identity, identical transformation, equivalent transformation, logical consequence.

The line of identical transformations is developed as follows in the course of algebra at the basic school:

, 4 b classes - opening brackets, bringing similar terms, take out- M (Chsho factor outside the brackets;

7 Class - identical transformations of integer and fractional expressions;

H class - identical transformations of expressions containing square-s roots;

( > class - identical transformations of trigonometric expressions and mmrizhsny, containing a degree with a rational exponent.

The line of identical transformations is one of the important ideological lines of the algebra course. Therefore, teaching mathematics in grades 5-6 is built in a way that students already in these grades acquire the skills of the simplest identical transformations (without using the term "identical transformations"). These skills are formed when performing an exercise on bringing similar terms, opening brackets and brackets, taking a factor out of brackets, etc. The simplest conversions of numeric and literal expressions are also considered. At this level of learning, transformations are mastered, which are performed directly on the basis of the laws and properties of arithmetic operations.

The main types of tasks in grades 5-6, in the solution of which the properties and laws of arithmetic operations are actively used and through which the skills of identical transformations are formed, include:

justification of algorithms for performing actions on the numbers of the studied numerical sets;

calculating the values ​​of a numeric expression in the most rational way;

comparison of values ​​of numeric expressions without performing the specified actions;

simplification of literal expressions;

proof of equality of the values ​​of two letter expressions, etc.

Present the number 153 as the sum of the digit terms; as a difference of two numbers, as a product of two numbers.

Imagine the number 27 as the product of three equal factors.

These exercises on the representation of the same number in different forms of notation contribute to the assimilation of the concept of identical transformations. Initially, these representations can be arbitrary, later on - purposeful. For example, the representation in the form of a sum of digit terms is used to explain the rules for adding natural numbers "in column", representation in the form of a sum or difference of "convenient" numbers - to perform quick calculations of various products, representation in the form of a product of factors - to simplify various fractional expressions.

Find the meaning of the expression 928 36 + 72 36.

The rational way to calculate the value of this expression is based on using the distribution law of multiplication relative to addition: 928 36 + 72 36 = (928 + 72) 36 = 1000 36 = 36000.

In the school course of mathematics, the following stages of mastering the applications of transformations of alphanumeric expressions and formulas can be distinguished.

stage. The beginnings of algebra. At this stage, an undivided system of transformations is used; it is represented by the rules for performing actions on one or both parts of the formula.

Example. Solve equations:

a) 5x - bx = 2; b) 5x = 3x + 2; v) 6 (2 - 4y) + 5y = 3 (1 - Zu).

The general idea behind the solution is to simplify these formulas with a few rules. In the first task simplification is achieved by applying the identity: 5x- Bx= (5 - 3) x. The identity transformation based on this identity transforms the given equation into an equivalent urshomie 2x - 2.

Second equation requires for its solution not only identical, but true transformation; in this capacity, the pra- || n is used here by transferring the terms of the equation from one part of the equation to another with a changed chic. In solving already such a simple task as b), both mon in transformations are used - both identical and equivalent. This provision is also true for more cumbersome tasks, such as the third.

The mole of the first stage is to teach how to quickly solve the simplest equations, to simplify formulas that define functions, to rationally carry out calculations based on the properties of actions.

tit. Formation of skills in the application of specific types of transformationsII tilt The concepts of identity and identical transformation are explicitly introduced in the course shn "sbry 7th grade. So, for example, in the textbook of Yu. N. Makarychev" Algebra 7 "nnp" shle, the concept of identically equal expressions is introduced: "Two expressions, the corresponding values ​​of which are equal for any values variables, sprinkle identically equal ", then the concept of identity: “Equality paired for any values ​​of variables is called identity ".

11 gives examples:

In the textbook A.G. Mordkovich's "Algebra 7" immediately gives a refined concept of identity: "Identity is equality true for any admissible the values ​​of its constituent variables ”.

When introducing the concept of identical transformations, one should first of all shake off the expediency of studying identical transformations. To do this, you can consider various exercises for finding the meaning of expressions.

liiiipiiMep, find the value of the expression 37.1x + 37, ly with X= 0.98, y = 0.02. Using the distributive property of multiplication, the expression 37.1l + 37.1 at can be expressed by the expression 37.1 (x + y), identically equal to him. Even more impressive worm 1 solution to the following exercise: find the meaning of the expression

() - (a-6) _ n p i. a) d = h> ^ = 2; b) a = 121, B - 38; c) a = 2.52, B = 1 -.

ab 9

11after the performed transformations, it turns out that the set of values ​​of this reflection consists of one number 4.

In the textbook "Algebra 7" by Yu. N. Makarychev, the introduction of the concept of an identical transformation is motivated by considering an example: "To find the meaning of the expression xy - yes at x = 2,3; y = 0.8; z = 0.2, you need to perform 3 actions: hu - xz = 2,3 0,8 - 2,3 0,2 = 1,84 - 0,46 = 1,38.

11 it is necessary to note one type of transformations specific to the course of algebra and the beginnings of analysis. These are transformations of expressions containing transitions, and transformations based on the rules of differentiation and integration. The main difference between these "analytic" transformations from "algebraic" transformations is in the character of the set, which runs through the variables in the identities. In algebraic identities, the variables run through number areas, and in analytic sets these sets ■ hang around certain many functions. For example, the rule of differential sum: (Z "+ g)" here / and g are variables running through the set

I I but differentiable functions with a common domain of definition. Outwardly, these transformations are similar to transformations of algebraic type, therefore sometimes they say "algebra of limits", "algebra of differentiation".

The identities studied in the school course of algebra and the algebraic ma-rial of the course of algebra and the principles of analysis can be divided into two classes.

The first consists of the abbreviated multiplication identities, fair in

av in.

iiioGom commutative ring, and the identities are = -, a * 0, which is valid in any

Oom field.

The second class is formed by identities connecting arithmetic numbers and basic elementary functions, as well as compositions of elementaryHhixfunctions. Most of the identities of this class also have a common mathematical basis, which is that the exponential, exponential, and logarithmic functions are isomorphisms of various numerical groups. For example, the following statement holds: there is a unique continuous isomorphic mapping / of the additive group of real numbers into the multiplicative group of positive real numbers, in which the unit is mapped to a given number a> 0, a f 1; this mapping is given by an incremental function with a radix a:/(X)= a. There are similar statements for power and logarithmic functions.

The methodology for studying identities in both classes has many common features. In general, the identical transformations studied in the school mathematics course include:

transformations of expressions containing radicals and powers with fractional exponents;

transformations of expressions containing limit transitions and transformations based on the rules of differentiation and integration.

This result can be obtained by performing only two steps - if you use the expression x (y-z), identically equal to the expression xy-xz: x (y-Z) = 2,3 (0,8 - 0,2) = 2,3 0,6 = 1,38.

We simplified the calculations by replacing the expression xy-xz identically equal expression x (y - z).

The replacement of one expression with another, identically equal to it is called identical transformation or simply by transforming an expression ".

Mastering various types of transformations at this stage begins with the introduction of abbreviated multiplication formulas. Then we consider transformations associated with the operation of raising to a power, with various classes of elementary functions - exponential, exponential, logarithmic, trigonometric. Each of these types of transformations goes through a stage of study, in which attention is focused on the assimilation of their characteristic features.

As the material accumulates, it becomes possible to single out and, on this basis, introduce the concepts of identical and equivalent transformations.

It should be noted that the concept of identical transformation is given in the school algebra course not in complete generality, but only in application to expressions. Transforms are divided into two classes: identical transformations are transformations of expressions, and equivalent - converting formulas. In the case when there is a need to simplify one part of the formula, an expression is highlighted in this formula, which serves as an argument for the applied identical transformation. For example, the equations 5x - Zx - 2 and 2x = 2 are considered not just equivalent, but the same.

In algebra textbooks Sh.A. Alimova et al., The concept of identity is not explicitly introduced in grades 7-8 and only in grade 9 in the topic "Trigonometric Identities" when solving problem 1: "Prove that for afkk, To < eZ , the equality 1 + ctg 2 a = - \ - is true, this concept is introduced. Here it is explained to the students that sin a

the indicated equality "is valid for all admissible values ​​of a, i.e. such that its left and right sides make sense. Such equalities are called identities, and problems of proving such equalities are called problems of proving identities. "

Stage III. Organization of an integral system of transformations (synthesis).

The main goal of this stage is to form a flexible and powerful apparatus suitable for use in solving a variety of educational tasks.

The deployment of the second stage of the study of transformations occurs throughout the entire course of basic school algebra. The transition to the third stage is carried out with the final repetition of the course in the course of comprehending the already known material, mastered in parts, for individual types of transformations.

In the course of algebra and the beginnings of analysis, the integral system of transformations, basically already formed, continues to gradually improve. Some new types of transformations are also added to it (for example, related to trigonometric and logarithmic functions), however, they only enrich it, expand its capabilities, but do not change its structure.

The methodology for studying these new transformations practically does not differ from that used in the algebra course.

It is necessary to note one type of transformations, specific to the kuren algebra and the beginnings of analysis. These are transformations of expressions containing limit transitions, and transformations based on the rules of differentiation and integration. The main difference between these "analytic" transformations and "algebraic" transformations is in the nature of the set that the variables run through in the identities. In algebraic identities, the variables run through number areas, and in analytic, these sets shine with certain many functions. For example, the rule for differentiating the amount: ( f + g )" = f + g "; here fug - variables running through multiple differentiable functions with a common domain of definition. Outwardly, these transformations are similar to transformations of algebraic type, therefore sometimes they say "algebra of limits", "algebra of differentiation".

The identities studied in the school course of algebra and the algebraic material of the course of algebra and the principles of analysis can be divided into two classes.

The first consists of the abbreviated multiplication identities, fair in

any commutative ring, and the identity - = -, a * 0, valid in any

ace with

The second class is formed by identities connecting arithmetic operations and basic elementary functions, as well as compositions of elementary functions. Most of the identities of this class also have a common mathematical basis, which is that the power, exponential, and logarithmic functions are isomorphisms of various numerical groups. For example, the following statement holds: there is a unique continuous isomorphic mapping / of the additive group of real numbers to the multiplicative group of positive real numbers, in which the unit is mapped to a given number a> 0, a f one; this mapping is given by an exponential function with radix i: / (x) = a *. There are similar statements for power and logarithmic functions.

The methodology for studying identities of both classes has many common features. In general, the identical transformations studied in the school mathematics course include:

transformations of algebraic expressions;

converting expressions containing radicals and powers with fractional exponents;

converting trigonometric expressions;

converting expressions containing degrees and logarithms;

transformations of expressions containing limit transitions and transformations based on rules, differentiation and integration.

2. Features of the organization of the system of tasks in the study of identical transformations

The basic principle of organizing any system of tasks is to present them from simple to complex taking into account the need for students to overcome feasible difficulties and create problem situations. This basic principle requires concretization in relation to the peculiarities of this educational material. Here is an example of a system of exercises on the topic: "The square of the sum and

difference of two numbers ".

I la this basic exercise system ends. Such a system should ensure the assimilation of the basic material.

The following exercises (17-19) allow students to focus on common mistakes and contribute to the development of interest and their creative 1 aids.

In each specific case, the number of exercises in the system can be less or more, but the sequence of their execution should be the same.

To describe various systems of tasks in the methodology of mathematics, the concept of exercise cycle. The cycle of exercises is characterized by the fact that several aspects of the study and techniques of the arrangement of the material are connected in a sequence of exercises. With respect to identical transformations, the concept of a cycle can be given as follows.

The 11th cycle of exercises is associated with the study of one identity, around which other identities are grouped that are in natural connection with it. In the "stop of the cycle along with executive includes tasks requiring recognizing< ii in nor the applicability of the identity under consideration. The studied identity is used to carry out calculations on various numerical areas.

The tasks in each cycle are divided into two groups. TO the first includes tasks performed at the initial acquaintance with identity. They are performed in several lessons, united by one topic. Second group exercises connects the studied identity with various applications. The exercises in this group are usually scattered across different topics.

The described structure of the cycle refers to the stage of the formation of skills in the application of specific types of transformations. At the final stage - (Tanya synthesis, the cycles are modified. Firstly, both groups of shdapiy are combined, forming Unrolled cycle , and from the first group are excluded the most simple in terms of wording or the complexity of the execution of the record. The remaining types of tasks get more complicated. Secondly, there is a merger of cycles related to different identities, due to this, the role of actions on recognizing the applicability of one or another identity increases.

11RNNSLet's give a concrete example of a loop.

Example. Cycle of tasks for identity x -y 2 = (x-y) (x + y).

The execution of the first group of tasks of this cycle occurs as follows -

conditions. The students have just familiarized themselves with the formulation of the identity (or rather, with two formulations: "The difference of the squares of two expressions is equal to the product of the sum and the difference of these expressions" and "The product of the sum and difference of two expressions is equal to the difference of the squares of these expressions"), its writing as a formula, proof ... After that, there are some examples of how to use a transformation based on this identity. Finally, students start doing the exercises on their own.

The first group of tasks

Second group of tasks

(The assignments of each group can be presented to students using a multimedia projector)

Let us carry out a methodological analysis of this system of task types.

The task a0 is intended to fix the structure of the identity under study. This is achieved by replacing the letters (x and y) in the notation of identity in other letters. Tasks of this type allow you to clarify the relationship between verbal expression and the symbolic form of identity.

Task a 2) is focused on establishing a connection between this identity and the number system. The expression to be converted is here not purely literal, but alphanumeric. To describe the actions performed, it is necessary to use the concept substitutions letters number in identity. Skills development

the application of the substitution operation and the deepening of the understanding of it carried out I um when performing tasks of the type d 2).

The next step in mastering the identity is illustrated by task a). In the nominal assignment, the expression proposed for transformation does not have the form of rasp n squares; transformation becomes possible only when. h (chp1k will notice that the number 121 can be represented as a square of a number. Thus, this task is performed not in one step, but in two: on the firstiiiiu there is a recognition of the possibility of reducing this expression to the MPD of the difference of squares, on the second a transformation is performed using the identity.

At the beginning of the development of identity, each step is recorded:

I "I / s 2 = 11 2 - & 2 = (11 - £) (11 + To), later, some recognition operations are performed by students orally.

In example dd), it is required to establish connections between this identity and others related to actions with monomials; in q 3) one should apply the identity for the difference of squares twice; c) students will have to overcome a certain psychological barrier, making their way into the area of ​​irrational numbers.

Tasks of type b) are aimed at developing skills for replacing the product (, v - y) (x + y) by the difference X 2 - at 2 . A similar role is played by tasks of type c). In examples of type d), it is required to choose one of the transformation directions.

In general, the tasks of the first group are focused on mastering the structure of the identity, substitution operations in the simplest most important cases, and ideas about the reversibility of transformations carried out by an identity,

The main features and goals, disclosed by us when considering the first | ruins of cycle assignments, refer to any cycle of exercise that forms the bayonets of the use of identity. For any newly introduced identity, the first group of tasks in the cycle must retain the features described here; differences can only be in the number of tasks.

1 The second group of tasks in the cycle, in contrast to the first, is aimed at the fullest possible use and taking into account the specifics of this particular identity, t i pi. Tasks of this group assume already formed skills of using the identity for the difference of squares (in the most simple cases); chi, the tasks of this group are to deepen the understanding of identity by considering its various applications in different situations, combined with the use of material related to other topics of the mathematics course.

Consider the solution to the task l):

x 3 - 4x = 15 o x 3 - 9x = 15 - 5x o x (x ~ 3) (x + 3) = 5 (3-x) x = 3, or \{\ 1-3) = -5. The equation x (x + 3) = -5 has no real roots, therefore \ 3 is the only root of the equation.

We see that the use of the identity for the difference of the squares is the pn and the I part in the solution of the example, being the leading idea of ​​carrying out the transformations.

Cycles of tasks associated with identities for elementary functions have their own characteristics, which are due to the fact that, in the first place... the corresponding identities are studied in connection with the study of functional material and, / u> - "toykh, they appear later than the identities of the first group and are studied with

using already formed skills of carrying out identical transformations. A significant part of the use of identical transformations associated with elementary functions falls on the solution of irrational and transcendental equations. The cycles related to the assimilation of identities include only the simplest equations, but already here it is advisable to work on mastering the technique of solving such equations: reducing it by replacing the unknown to an algebraic equation.

The sequence of steps for this solution is as follows:

a) find the function<р, для которой данное уравнение/(х) = 0 представимо в виде F (ср(лг)) = 0;

b) make substitution at= cp (x) and solve the equation F (y) = 0;

c) solve each of the equations <р(х) = where (at j) is the set of roots of the equation F (y) = 0.

A new issue that must be taken into account when studying identities with elementary functions is the consideration of the domain of definition. Here are examples of three tasks:

a) Plot the function y = 4 log 2 x.

b) Solve the equation lg X + lg (x - 3) = 1.

c) On what set is the formula lg (x - 5) + lg (x + 5) = lg ( X 2 - 25) is an identity?

A typical mistake that students make in solving task a) is to use equality a 1st excluding condition B> 0. In this case, as a result, the desired graph turns out to have the form of a parabola instead of the correct answer - the right branch of the parabola. In task b) one of the sources for obtaining complex systems of equations and inequalities is shown, when it is necessary to take into account the domains of definition of functions, and in task c) - an exercise that can serve as a preparatory one.

The idea that unites these tasks - the need to study the area of ​​definition of the function, can only come to light when comparing such tasks, dissimilar in their external form. The significance of this idea for mathematics is very great. It can serve as the basis for several cycles of exercises - for each of the classes of elementary functions.

In conclusion, we note that the study of identical transformations in school has a large educational value. The ability to do some calculations, carry out calculations, for a long time with unremitting attention to follow some object is necessary for people of a wide variety of professions, regardless of whether they work in the field of mental or physical labor. The specificity of the section "Identical transformations of expressions" is such that it opens up ample opportunities for students to develop these important professionally significant skills.

The numbers and expressions from which the original expression is composed can be replaced by identically equal expressions. Such a transformation of the original expression leads to an expression identically equal to it.

For example, in the expression 3 + x, the number 3 can be replaced with the sum 1 + 2, and the expression (1 + 2) + x will be obtained, which is identically equal to the original expression. Another example: in the expression 1 + a 5, the degree of a 5 can be replaced by an identically equal product, for example, of the form a · a 4. This will give us the expression 1 + a · a 4.

This transformation is undoubtedly artificial and usually prepares for some further transformation. For example, in the sum 4 · x 3 + 2 · x 2, taking into account the properties of the degree, the term 4 · x 3 can be represented as the product 2 · x 2 · 2 · x. After this transformation, the original expression will take the form 2 · x 2 · 2 · x + 2 · x 2. Obviously, the terms in the resulting sum have a common factor 2 · x 2, so we can perform the following transformation - parentheses. After that, we come to the expression: 2 x 2 (2 x + 1).

Add and subtract the same number

Another artificial transformation of an expression is the addition and subtraction of the same number or expression at the same time. This conversion is identical, since it is essentially equivalent to adding zero, and adding zero does not change the value.

Let's look at an example. Take the expression x 2 + 2 x. If we add one to it and subtract one, then this will allow us to perform one more identical transformation in the future - select the square of the binomial: x 2 + 2 x = x 2 + 2 x + 1−1 = (x + 1) 2 −1.

Bibliography.

• Algebra: study. for 7 cl. general education. institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 17th ed. - M.: Education, 2008 .-- 240 p. : ill. - ISBN 978-5-09-019315-3.
• Algebra: study. for 8 cl. general education. institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M.: Education, 2008 .-- 271 p. : ill. - ISBN 978-5-09-019243-9.
• A. G. Mordkovich Algebra. 7th grade. At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich. - 17th ed., Add. - M .: Mnemozina, 2013 .-- 175 p .: ill. ISBN 978-5-346-02432-3.

Lesson content

Exponentiation of a binomial

A binomial is a polynomial with two members. 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 a binomial can be raised not only to the second and third degrees, but also to the fourth, fifth, or higher degrees.

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

Cofactor ( a + b) 3 can be replaced with the right side of the cube formula for 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 the fourth degree is a polynomial 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

Erection of a binomial a + b to the fourth degree, you can also perform this way: represent the expression ( a + b) 4 as a product of degrees (a + b) 2 (a + b) 2

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

But the expression ( a + b) 2 equals a 2 + 2ab + b 2 ... Replace in the expression (a + b) 2 (a + b) 2 the squares of the sum by the polynomial a 2 + 2ab + b 2

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

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

Exponentiation of a trinomial

A three-term is a polynomial with three members. For example, the expression a + b + c is a three-term.

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

(a + b + c) 2

Two terms within parentheses can be enclosed in parentheses. For example, let us 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 not squaring a three-term, but a two-term. 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, 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, we enclose them in brackets:

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

Now let's use the formula for the square of the sum of two expressions:

Isolation of a complete square from a square trinomial

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

A square trinomial is a second degree trinomial. For example, the following three terms are square:

The idea of ​​isolating a complete square from such trinomials is to represent the original square trinomial in the form of the expression ( a + b) 2 + c, where ( a + b) 2 is a complete square, and c - some numeric or literal expression.

For example, let's select a complete square from a 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 define which members will play the role of variables a and b

The role of a variable a will play member 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 is equal to 2 x

a = 2x

Now we return to the original three-term and immediately pay attention to the expression 16 x... This expression is the double product of the first expression a(in our case it is 2 x) and the second expression still unknown to us b. Let's temporarily put a question mark in its place:

2 × 2 x × ? = 16x

If you look closely at the expression 2 × 2 x × ? = 16x , then it becomes intuitively clear that the term b in this situation, the number 4 is because the expression 2 × 2 x equals 4 x, and to get 16 x you need to multiply 4 x by 4.

2 × 2 x × 4 = 16x

Hence, we conclude that the variable b equals 4

b = 4

This means that our complete square will be the expression (2x) 2 + 2 × 2 x× 4 + 4 2

Now we are ready to select a complete square from a trinomial. 4x 2 + 16x+ 19 .

So, back to the original trinomial 4x 2 + 16x+ 19 and we will try to carefully introduce the complete square we received into it (2x) 2 + 2 × 2 x× 4 + 4 2

4x 2 + 16x+ 19 =

Instead of 4 x 2 we write (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 while we rewrite term 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 obtained polynomial (2x) 2 + 2 × 2 x× 4 + 4 2 + 19 not identical to the original three-term 4x 2 + 16x+ 19 ... You can verify this by bringing the polynomial (2x) 2 + 2 × 2 x× 4 + 4 2 + 19 to the 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 the term 4 2 was artificially implanted into the original tri-term in order to organize a complete square of the tri-term 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 complete 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 a 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 doubled product of the first expression (we have x) and the second expression b(this will be 1).

2 × x× 1 = 2 x

If b= 1, then the expression x 2 + 2x+ 1 2 .

Now let's go back to the original square trinomial and embed a complete square in 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, this is 1) after the addition, it was immediately subtracted 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 complete square in this numerical expression. To do this, we first construct an expression of the form a 2 + 2ab+ b 2 ... The role of a 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.

The second term 6 is represented as the doubled product of the first term 3 and the second 1

2 × 3 × 1 = 6

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

− 1 2 + 2

Let's fold a complete square, and the terms −1 2 and 2 can be added:

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

The result is the expression (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

Let's calculate the area of ​​each shape. The area of ​​the square will be 3 2 = 9 cm 2, the area of ​​the pink rectangle - 2 × 3 = 6 cm 2, the area of ​​the lilac - 1 × 2 = 2 cm 2

Let's 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 shape:

Then a figure is obtained, the area of ​​which is 17 cm 2. Indeed, the figure shown contains 17 squares with a side of 1 cm.

Let's try to form a square from the existing figure. Moreover, the largest square. For this we will use parts from the pink and purple rectangle.

To form the largest square from the existing figure, 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, take one square from the lilac rectangle and attach it to the large square that is formed:

Now let's take a close look at what we have come to. Namely, on the yellow part of the figure and the pink part, which in fact increased the previous yellow square. Does 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 the area?

(3 + 1) 2

Expression (3 + 1) 2 is 16 because 3 + 1 = 4 and 4 2 = 16. The same result can be obtained by 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 purple 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 a complete 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 17. Indeed, the area of ​​the formed figure is 17 cm 2.

Example 4... Let's perform the selection of a complete square from a 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 select a complete square from a square trinomial x 2 + 3x+ 2

Variable a corresponds to x... Second term 3 x cannot be represented as a doubled 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.

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Slide captions:

Identities. Identical 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. From the distribution property it follows 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

Let us now consider 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 for x = 3, y = 4 the values ​​of the expressions are different 2x + y = 2 * 3 + 4 = 10 2xy = 2 * 3 * 4 = 24

CONCLUSION: Expressions 3 (x + y) and 3x + 3y are identically equal, but expressions 2x + y and 2xy are not identically equal. Definition: Two expressions, the values ​​of which are equal for any values ​​of the variables, are called 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: Equality, true for any values ​​of the variables, is called 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. a + b = b + a ab = ba (a + b) + c = a + (b + c) (ab) c = a (bc) a (b + c) = ab + ac

You can give other examples of identities: a + 0 = a a * 1 = a a + (-a) = 0 a * (- b) = - ab a- b = a + (- b) (-a) * ( -b) = ab Replacing one expression with another, identically equal expression, is called an identity conversion, or simply an expression conversion.

To give such terms, you need to add their coefficients and multiply the result by the total letter part. Example 1. Let's give similar terms 5x + 2x-3x = x (5 + 2-3) = 4x

If there is a plus sign in front of the brackets, then the brackets can be omitted, keeping the sign of each term enclosed in brackets. Example 2. Let's expand the brackets in the expression 2а + (b -3 c) = 2 a + b - 3 c

If there is a minus sign in front of 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) = a - 4 b + c

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

On the subject: methodological developments, presentations and notes

Methodology for preparing students for the Unified State Exam in the section "Expressions and Expression Transformation"

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