Expressions, expression conversion

Power expressions (expressions with powers) and their transformation

In this article we will talk about converting expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening parentheses and bringing similar terms. And then we will analyze the transformations inherent specifically in expressions with degrees: working with the base and exponent, using the properties of degrees, etc.

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What are power expressions?

The term " power expressions"is practically not found in school mathematics textbooks, but it appears quite often in collections of problems, especially those intended for preparation for the Unified State Exam and the Unified State Exam, for example. After analyzing the tasks in which it is necessary to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing powers in their entries. Therefore, you can accept the following definition for yourself:

Definition.

Power expressions are expressions containing powers.

Let's give examples of power expressions. Moreover, we will present them according to how the development of views on from degree to degree occurs. natural indicator to a degree with a real exponent.

As is known, first one gets acquainted with the power of a number with a natural exponent; at this stage, the first simplest power expressions of the type 3 2, 7 5 +1, (2+1) 5, (−0.1) 4, 3 a 2 appear −a+a 2 , x 3−1 , (a 2) 3 etc.

A little later, the power of a number with an integer exponent is studied, which leads to the appearance of power expressions with integers negative powers, like the following: 3 −2 , , a −2 +2 b −3 +c 2 .

In high school they return to degrees. There a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: , , and so on. Finally, degrees with irrational exponents and expressions containing them are considered: , .

The matter is not limited to the listed power expressions: further the variable penetrates into the exponent, and, for example, the following expressions arise: 2 x 2 +1 or . And after getting acquainted with , expressions with powers and logarithms begin to appear, for example, x 2·lgx −5·x lgx.

So, we have dealt with the question of what power expressions represent. Next we will learn to convert them.

Main types of transformations of power expressions

With power expressions, you can perform any of the basic identity transformations of expressions. For example, you can open parentheses, replace numerical expressions with their values, add similar terms, etc. Naturally, in this case, it is necessary to follow the accepted procedure for performing actions. Let's give examples.

Example.

Calculate the value of the power expression 2 3 ·(4 2 −12) .

Solution.

According to the order of execution of actions, first perform the actions in brackets. There, firstly, we replace the power 4 2 with its value 16 (if necessary, see), and secondly, we calculate the difference 16−12=4. We have 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4.

In the resulting expression, we replace the power 2 3 with its value 8, after which we calculate the product 8·4=32. This is the desired value.

So, 2 3 ·(4 2 −12)=2 3 ·(16−12)=2 3 ·4=8·4=32.

Answer:

2 3 ·(4 2 −12)=32.

Example.

Simplify expressions with powers 3 a 4 b −7 −1+2 a 4 b −7.

Solution.

Obviously, this expression contains similar terms 3·a 4 ·b −7 and 2·a 4 ·b −7 , and we can present them: .

Answer:

3 a 4 b −7 −1+2 a 4 b −7 =5 a 4 b −7 −1.

Example.

Express an expression with powers as a product.

Solution.

You can cope with the task by representing the number 9 as a power of 3 2 and then using the formula for abbreviated multiplication - difference of squares:

Answer:

There are also a number identity transformations, inherent specifically in power expressions. We will analyze them further.

Working with base and exponent

There are degrees whose base and/or exponent are not just numbers or variables, but some expressions. As an example, we give the entries (2+0.3·7) 5−3.7 and (a·(a+1)−a 2) 2·(x+1) .

When working with such expressions, you can replace both the expression in the base of the degree and the expression in the exponent with an identically equal expression in the ODZ of its variables. In other words, according to the rules known to us, we can separately transform the base of the degree and separately the exponent. It is clear that as a result of this transformation, an expression will be obtained that is identically equal to the original one.

Such transformations allow us to simplify expressions with powers or achieve other goals we need. For example, in the power expression mentioned above (2+0.3 7) 5−3.7, you can perform operations with the numbers in the base and exponent, which will allow you to move to the power 4.1 1.3. And after opening the brackets and bringing similar terms to the base of the degree (a·(a+1)−a 2) 2·(x+1) we obtain a power expression more simple type a 2·(x+1) .

Using Degree Properties

One of the main tools for transforming expressions with powers is equalities that reflect . Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following properties of powers are true:

• a r ·a s =a r+s ;
• a r:a s =a r−s ;
• (a·b) r =a r ·b r ;
• (a:b) r =a r:b r ;
• (a r) s =a r·s .

Note that for natural, integer, and positive exponents, the restrictions on the numbers a and b may not be so strict. For example, for natural numbers m and n the equality a m ·a n =a m+n is true not only for positive a, but also for negative a, and for a=0.

At school, the main focus when transforming power expressions is on the ability to choose the appropriate property and apply it correctly. In this case, the bases of degrees are usually positive, which allows the properties of degrees to be used without restrictions. The same applies to the transformation of expressions containing variables in the bases of powers - the range of permissible values ​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of powers. In general, you need to constantly ask yourself whether it is possible to use any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the educational value and other troubles. These points are discussed in detail and with examples in the article transformation of expressions using properties of degrees. Here we will limit ourselves to considering a few simple examples.

Example.

Express the expression a 2.5 ·(a 2) −3:a −5.5 as a power with base a.

Solution.

First, we transform the second factor (a 2) −3 using the property of raising a power to a power: (a 2) −3 =a 2·(−3) =a −6. The original power expression will take the form a 2.5 ·a −6:a −5.5. Obviously, it remains to use the properties of multiplication and division of powers with the same base, we have
a 2.5 ·a −6:a −5.5 =
a 2.5−6:a −5.5 =a −3.5:a −5.5 =
a −3.5−(−5.5) =a 2 .

Answer:

a 2.5 ·(a 2) −3:a −5.5 =a 2.

Properties of powers when transforming power expressions are used both from left to right and from right to left.

Example.

Find the value of the power expression.

Solution.

The equality (a·b) r =a r ·b r, applied from right to left, allows us to move from the original expression to a product of the form and further. And when multiplying powers with on the same grounds the indicators add up: .

It was possible to transform the original expression in another way:

Answer:

.

Example.

Given the power expression a 1.5 −a 0.5 −6, introduce a new variable t=a 0.5.

Solution.

The degree a 1.5 can be represented as a 0.5 3 and then, based on the property of the degree to the degree (a r) s =a r s, applied from right to left, transform it to the form (a 0.5) 3. Thus, a 1.5 −a 0.5 −6=(a 0.5) 3 −a 0.5 −6. Now it’s easy to introduce a new variable t=a 0.5, we get t 3 −t−6.

Answer:

t 3 −t−6 .

Converting fractions containing powers

Power expressions can contain or represent fractions with powers. Any of the basic transformations of fractions that are inherent in fractions of any kind are fully applicable to such fractions. That is, fractions that contain powers can be reduced, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate these words, consider solutions to several examples.

Example.

Simplify power expression .

Solution.

This power expression is a fraction. Let's work with its numerator and denominator. In the numerator we open the brackets and simplify the resulting expression using the properties of powers, and in the denominator we present similar terms:

And let’s also change the sign of the denominator by placing a minus in front of the fraction: .

Answer:

.

Reducing fractions containing powers to a new denominator is carried out similarly to reducing rational fractions to a new denominator. In this case, an additional factor is also found and the numerator and denominator of the fraction are multiplied by it. When performing this action, it is worth remembering that reduction to a new denominator can lead to a narrowing of the VA. To prevent this from happening, it is necessary that the additional factor does not go to zero for any values ​​of the variables from the ODZ variables for the original expression.

Example.

Reduce the fractions to a new denominator: a) to denominator a, b) to the denominator.

Solution.

a) In this case, it is quite easy to figure out what additional multiplier helps to achieve desired result. This is a multiplier of a 0.3, since a 0.7 ·a 0.3 =a 0.7+0.3 =a. Note that in the range of permissible values ​​of the variable a (this is the set of all positive real numbers), the power of a 0.3 does not vanish, therefore, we have the right to multiply the numerator and denominator of a given fraction by this additional factor:

b) Taking a closer look at the denominator, you will find that

and multiplying this expression by will give the sum of cubes and , that is, . And this is the new denominator to which we need to reduce the original fraction.

This is how we found an additional factor. In the range of permissible values ​​of the variables x and y, the expression does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:

Answer:

A) , b) .

There is also nothing new in reducing fractions containing powers: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are reduced.

Example.

Reduce the fraction: a) , b) .

Solution.

a) Firstly, the numerator and denominator can be reduced by the numbers 30 and 45, which is equal to 15. It is also obviously possible to perform a reduction by x 0.5 +1 and by . Here's what we have:

b) In this case, identical factors in the numerator and denominator are not immediately visible. To obtain them, you will have to perform preliminary transformations. In this case, they consist in factoring the denominator using the difference of squares formula:

Answer:

A)

b) .

Converting fractions to a new denominator and reducing fractions are mainly used to do things with fractions. Actions are performed according to known rules. When adding (subtracting) fractions, they are reduced to a common denominator, after which the numerators are added (subtracted), but the denominator remains the same. The result is a fraction whose numerator is the product of the numerators, and the denominator is the product of the denominators. Division by a fraction is multiplication by its inverse.

Example.

Follow the steps .

Solution.

First, we subtract the fractions in parentheses. To do this, we bring them to a common denominator, which is , after which we subtract the numerators:

Now we multiply the fractions:

Obviously, it is possible to reduce by a power of x 1/2, after which we have .

You can also simplify the power expression in the denominator by using the difference of squares formula: .

Answer:

Example.

Simplify the Power Expression .

Solution.

Obviously, this fraction can be reduced by (x 2.7 +1) 2, this gives the fraction . It is clear that something else needs to be done with the powers of X. To do this, we transform the resulting fraction into a product. This gives us the opportunity to take advantage of the property of dividing powers with the same bases: . And at the end of the process we move from the last product to the fraction.

Answer:

.

And let us also add that it is possible, and in many cases desirable, to transfer factors with negative exponents from the numerator to the denominator or from the denominator to the numerator, changing the sign of the exponent. Such transformations often simplify further actions. For example, a power expression can be replaced by .

Converting expressions with roots and powers

Often, in expressions in which some transformations are required, roots with fractional exponents are also present along with powers. To convert such an expression to the right type, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with powers, they usually move from roots to powers. However, it is advisable to carry out such a transition when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to refer to the module or split the ODZ into several intervals (we discussed this in detail in the article transition from roots to powers and back After getting acquainted with the degree with a rational exponent a degree with an irrational exponent is introduced, which allows us to talk about a degree with an arbitrary real exponent. At this stage, the school begins to study exponential function, which is analytically given by a power, the base of which is a number, and the exponent is a variable. So we are faced with power expressions containing numbers in the base of the power, and in the exponent - expressions with variables, and naturally the need arises to perform transformations of such expressions.

It should be said that the transformation of expressions of the indicated type usually has to be performed when solving exponential equations And exponential inequalities, and these conversions are quite simple. In the overwhelming majority of cases, they are based on the properties of the degree and are aimed, for the most part, at introducing a new variable in the future. The equation will allow us to demonstrate them 5 2 x+1 −3 5 x 7 x −14 7 2 x−1 =0.

Firstly, powers, in the exponents of which is the sum of a certain variable (or expression with variables) and a number, are replaced by products. This applies to the first and last terms of the expression on the left side:
5 2 x 5 1 −3 5 x 7 x −14 7 2 x 7 −1 =0,
5 5 2 x −3 5 x 7 x −2 7 2 x =0.

Next, both sides of the equality are divided by the expression 7 2 x, which on the ODZ of the variable x for the original equation takes only positive values ​​(this is a standard technique for solving equations of this type, we are not talking about it now, so focus on subsequent transformations of expressions with powers ):

Now we can cancel fractions with powers, which gives .

Finally, the ratio of powers with the same exponents is replaced by powers of relations, resulting in the equation , which is equivalent . The transformations made allow us to introduce a new variable, which reduces the solution to the original exponential equation to solving a quadratic equation

I. V. Boykov, L. D. Romanova Collection of tasks for preparing for the Unified State Exam. Part 1. Penza 2003.

Let's consider the topic of transforming expressions with powers, but first let's dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open parentheses, add similar terms, work with bases and exponents, and use the properties of powers.

Yandex.RTB R-A-339285-1

What are power expressions?

IN school course Few people use the phrase “power expressions,” but this term is constantly found in collections for preparing for the Unified State Exam. In most cases, a phrase denotes expressions that contain degrees in their entries. This is what we will reflect in our definition.

Definition 1

Power expression is an expression that contains powers.

Let us give several examples of power expressions, starting with a power with a natural exponent and ending with a power with a real exponent.

The simplest power expressions can be considered powers of a number with a natural exponent: 3 2, 7 5 + 1, (2 + 1) 5, (− 0, 1) 4, 2 2 3 3, 3 a 2 − a + a 2, x 3 − 1 , (a 2) 3 . And also powers with zero exponent: 5 0, (a + 1) 0, 3 + 5 2 − 3, 2 0. And powers with negative integer powers: (0, 5) 2 + (0, 5) - 2 2.

It is a little more difficult to work with a degree that has rational and irrational exponents: 264 1 4 - 3 3 3 1 2, 2 3, 5 2 - 2 2 - 1, 5, 1 a 1 4 a 1 2 - 2 a - 1 6 · b 1 2 , x π · x 1 - π , 2 3 3 + 5 .

The indicator can be the variable 3 x - 54 - 7 3 x - 58 or the logarithm x 2 · l g x − 5 · x l g x.

We have dealt with the question of what power expressions are. Now let's start converting them.

Main types of transformations of power expressions

First of all, we will look at the basic identity transformations of expressions that can be performed with power expressions.

Example 1

Calculate the value of a power expression 2 3 (4 2 − 12).

Solution

We will carry out all transformations in compliance with the order of actions. In this case, we will start by performing the actions in brackets: we will replace the degree with a digital value and calculate the difference of two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.

All we have to do is replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here's our answer.

Answer: 2 3 · (4 2 − 12) = 32 .

Example 2

Simplify the expression with powers 3 a 4 b − 7 − 1 + 2 a 4 b − 7.

Solution

The expression given to us in the problem statement contains similar terms that we can give: 3 a 4 b − 7 − 1 + 2 a 4 b − 7 = 5 a 4 b − 7 − 1.

Answer: 3 · a 4 · b − 7 − 1 + 2 · a 4 · b − 7 = 5 · a 4 · b − 7 − 1 .

Example 3

Express the expression with powers 9 - b 3 · π - 1 2 as a product.

Solution

Let's imagine the number 9 as a power 3 2 and apply the abbreviated multiplication formula:

9 - b 3 π - 1 2 = 3 2 - b 3 π - 1 2 = = 3 - b 3 π - 1 3 + b 3 π - 1

Answer: 9 - b 3 · π - 1 2 = 3 - b 3 · π - 1 3 + b 3 · π - 1 .

Now let's move on to the analysis of identity transformations that can be applied specifically to power expressions.

Working with base and exponent

The degree in the base or exponent can have numbers, variables, and some expressions. For example, (2 + 0, 3 7) 5 − 3, 7 And . Working with such records is difficult. It is much easier to replace the expression in the base of the degree or the expression in the exponent with an identically equal expression.

Transformations of degree and exponent are carried out according to the rules known to us separately from each other. The most important thing is that the transformation results in an expression identical to the original one.

The purpose of transformations is to simplify the original expression or obtain a solution to the problem. For example, in the example we gave above, (2 + 0, 3 7) 5 − 3, 7 you can follow the steps to go to the degree 4 , 1 1 , 3 . By opening the parentheses, we can present similar terms to the base of the power (a · (a + 1) − a 2) 2 · (x + 1) and obtain a power expression of a simpler form a 2 (x + 1).

Using Degree Properties

Properties of powers, written in the form of equalities, are one of the main tools for transforming expressions with powers. We present here the main ones, taking into account that a And b are any positive numbers, and r And s- arbitrary real numbers:

Definition 2

• a r · a s = a r + s ;
• a r: a s = a r − s ;
• (a · b) r = a r · b r ;
• (a: b) r = a r: b r ;
• (a r) s = a r · s .

In cases where we are dealing with natural, integer, positive exponents, the restrictions on the numbers a and b can be much less strict. So, for example, if we consider the equality a m · a n = a m + n, Where m And n are natural numbers, then it will be true for any values ​​of a, both positive and negative, as well as for a = 0.

The properties of powers can be used without restrictions in cases where the bases of the powers are positive or contain variables whose range of permissible values ​​is such that the bases take only positive values ​​on it. In fact, in the school mathematics curriculum, the student's task is to select an appropriate property and apply it correctly.

When preparing to enter universities, you may encounter problems in which inaccurate application of properties will lead to a narrowing of the DL and other difficulties in solving. In this section we will examine only two such cases. More information on the subject can be found in the topic “Converting expressions using properties of powers”.

Example 4

Imagine the expression a 2 , 5 (a 2) − 3: a − 5 , 5 in the form of a power with a base a.

Solution

First, we use the property of exponentiation and transform the second factor using it (a 2) − 3. Then we use the properties of multiplication and division of powers with the same base:

a 2 , 5 · a − 6: a − 5 , 5 = a 2 , 5 − 6: a − 5 , 5 = a − 3 , 5: a − 5 , 5 = a − 3 , 5 − (− 5 , 5) = a 2 .

Answer: a 2, 5 · (a 2) − 3: a − 5, 5 = a 2.

Transformation of power expressions according to the property of powers can be done both from left to right and in the opposite direction.

Example 5

Find the value of the power expression 3 1 3 · 7 1 3 · 21 2 3 .

Solution

If we apply equality (a · b) r = a r · b r, from right to left, we get a product of the form 3 · 7 1 3 · 21 2 3 and then 21 1 3 · 21 2 3 . Let's add the exponents when multiplying powers with the same bases: 21 1 3 · 21 2 3 = 21 1 3 + 2 3 = 21 1 = 21.

There is another way to carry out the transformation:

3 1 3 · 7 1 3 · 21 2 3 = 3 1 3 · 7 1 3 · (3 · 7) 2 3 = 3 1 3 · 7 1 3 · 3 2 3 · 7 2 3 = = 3 1 3 · 3 2 3 7 1 3 7 2 3 = 3 1 3 + 2 3 7 1 3 + 2 3 = 3 1 7 1 = 21

Answer: 3 1 3 7 1 3 21 2 3 = 3 1 7 1 = 21

Example 6

Given a power expression a 1, 5 − a 0, 5 − 6, enter a new variable t = a 0.5.

Solution

Let's imagine the degree a 1, 5 How a 0.5 3. Using the property of degrees to degrees (a r) s = a r · s from right to left and we get (a 0, 5) 3: a 1, 5 − a 0, 5 − 6 = (a 0, 5) 3 − a 0, 5 − 6. You can easily introduce a new variable into the resulting expression t = a 0.5: we get t 3 − t − 6.

Answer: t 3 − t − 6 .

Converting fractions containing powers

We usually deal with two versions of power expressions with fractions: the expression represents a fraction with a power or contains such a fraction. All basic transformations of fractions are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, or worked separately with the numerator and denominator. Let's illustrate this with examples.

Example 7

Simplify the power expression 3 · 5 2 3 · 5 1 3 - 5 - 2 3 1 + 2 · x 2 - 3 - 3 · x 2 .

Solution

We are dealing with a fraction, so we will carry out transformations in both the numerator and the denominator:

3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = 3 5 2 3 5 1 3 - 3 5 2 3 5 - 2 3 - 2 - x 2 = = 3 5 2 3 + 1 3 - 3 5 2 3 + - 2 3 - 2 - x 2 = 3 5 1 - 3 5 0 - 2 - x 2

Place a minus sign in front of the fraction to change the sign of the denominator: 12 - 2 - x 2 = - 12 2 + x 2

Answer: 3 5 2 3 5 1 3 - 5 - 2 3 1 + 2 x 2 - 3 - 3 x 2 = - 12 2 + x 2

Fractions containing powers are reduced to a new denominator in the same way as rational fractions. To do this, you need to find an additional factor and multiply the numerator and denominator of the fraction by it. It is necessary to select an additional factor in such a way that it does not go to zero for any values ​​of variables from the ODZ variables for the original expression.

Example 8

Reduce the fractions to a new denominator: a) a + 1 a 0, 7 to the denominator a, b) 1 x 2 3 - 2 · x 1 3 · y 1 6 + 4 · y 1 3 to the denominator x + 8 · y 1 2 .

Solution

a) Let's select a factor that will allow us to reduce to a new denominator. a 0, 7 a 0, 3 = a 0, 7 + 0, 3 = a, therefore, as an additional factor we will take a 0 , 3. The range of permissible values ​​of the variable a includes the set of all positive real numbers. Degree in this field a 0 , 3 does not go to zero.

Let's multiply the numerator and denominator of a fraction by a 0 , 3:

a + 1 a 0, 7 = a + 1 a 0, 3 a 0, 7 a 0, 3 = a + 1 a 0, 3 a

b) Let's pay attention to the denominator:

x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 2 - x 1 3 2 y 1 6 + 2 y 1 6 2

Let's multiply this expression by x 1 3 + 2 · y 1 6, we get the sum of the cubes x 1 3 and 2 · y 1 6, i.e. x + 8 · y 1 2 . This is our new denominator to which we need to reduce the original fraction.

This is how we found the additional factor x 1 3 + 2 · y 1 6 . On the range of permissible values ​​of variables x And y the expression x 1 3 + 2 y 1 6 does not vanish, therefore, we can multiply the numerator and denominator of the fraction by it:
1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 + 2 y 1 6 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = = x 1 3 + 2 y 1 6 x 1 3 3 + 2 y 1 6 3 = x 1 3 + 2 y 1 6 x + 8 y 1 2

Answer: a) a + 1 a 0, 7 = a + 1 a 0, 3 a, b) 1 x 2 3 - 2 x 1 3 y 1 6 + 4 y 1 3 = x 1 3 + 2 y 1 6 x + 8 · y 1 2 .

Example 9

Reduce the fraction: a) 30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3, b) a 1 4 - b 1 4 a 1 2 - b 1 2.

Solution

a) We use the greatest common denominator (GCD), by which we can reduce the numerator and denominator. For numbers 30 and 45 it is 15. We can also make a reduction by x0.5+1 and on x + 2 · x 1 1 3 - 5 3 .

We get:

30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 x 3 3 (x 0, 5 + 1)

b) Here the presence of identical factors is not obvious. You will have to perform some transformations in order to get the same factors in the numerator and denominator. To do this, we expand the denominator using the difference of squares formula:

a 1 4 - b 1 4 a 1 2 - b 1 2 = a 1 4 - b 1 4 a 1 4 2 - b 1 2 2 = = a 1 4 - b 1 4 a 1 4 + b 1 4 a 1 4 - b 1 4 = 1 a 1 4 + b 1 4

Answer: a) 30 x 3 (x 0, 5 + 1) x + 2 x 1 1 3 - 5 3 45 x 0, 5 + 1 2 x + 2 x 1 1 3 - 5 3 = 2 · x 3 3 · (x 0 , 5 + 1) , b) a 1 4 - b 1 4 a 1 2 - b 1 2 = 1 a 1 4 + b 1 4 .

Basic operations with fractions include converting fractions to a new denominator and reducing fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, first the fractions are reduced to a common denominator, after which operations (addition or subtraction) are carried out with the numerators. The denominator remains the same. The result of our actions is a new fraction, the numerator of which is the product of the numerators, and the denominator is the product of the denominators.

Example 10

Do the steps x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 · 1 x 1 2 .

Solution

Let's start by subtracting the fractions that are in parentheses. Let's bring them to a common denominator:

x 1 2 - 1 x 1 2 + 1

Let's subtract the numerators:

x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 + 1 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 - x 1 2 - 1 x 1 2 - 1 x 1 2 + 1 x 1 2 - 1 1 x 1 2 = = x 1 2 + 1 2 - x 1 2 - 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = x 1 2 2 + 2 x 1 2 + 1 - x 1 2 2 - 2 x 1 2 + 1 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2

Now we multiply the fractions:

4 x 1 2 x 1 2 - 1 x 1 2 + 1 1 x 1 2 = = 4 x 1 2 x 1 2 - 1 x 1 2 + 1 x 1 2

Let's reduce by a power x 1 2, we get 4 x 1 2 - 1 · x 1 2 + 1 .

Additionally, you can simplify the power expression in the denominator using the difference of squares formula: squares: 4 x 1 2 - 1 x 1 2 + 1 = 4 x 1 2 2 - 1 2 = 4 x - 1 .

Answer: x 1 2 + 1 x 1 2 - 1 - x 1 2 - 1 x 1 2 + 1 1 x 1 2 = 4 x - 1

Example 11

Simplify the power-law expression x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3.
Solution

We can reduce the fraction by (x 2 , 7 + 1) 2. We get the fraction x 3 4 x - 5 8 x 2, 7 + 1.

Let's continue transforming the powers of x x 3 4 x - 5 8 · 1 x 2, 7 + 1. Now you can use the property of dividing powers with the same bases: x 3 4 x - 5 8 1 x 2, 7 + 1 = x 3 4 - - 5 8 1 x 2, 7 + 1 = x 1 1 8 1 x 2 , 7 + 1 .

We move from the last product to the fraction x 1 3 8 x 2, 7 + 1.

Answer: x 3 4 x 2, 7 + 1 2 x - 5 8 x 2, 7 + 1 3 = x 1 3 8 x 2, 7 + 1.

In most cases, it is more convenient to transfer factors with negative exponents from the numerator to the denominator and back, changing the sign of the exponent. This action allows you to simplify the further decision. Let's give an example: the power expression (x + 1) - 0, 2 3 · x - 1 can be replaced by x 3 · (x + 1) 0, 2.

Converting expressions with roots and powers

In problems there are power expressions that contain not only powers with fractional exponents, but also roots. It is advisable to reduce such expressions only to roots or only to powers. Going for degrees is preferable as they are easier to work with. This transition is especially preferable when the ODZ of variables for the original expression allows you to replace the roots with powers without the need to access the modulus or split the ODZ into several intervals.

Example 12

Express the expression x 1 9 · x · x 3 6 as a power.

Solution

Range of permissible variable values x is defined by two inequalities x ≥ 0 and x x 3 ≥ 0, which define the set [ 0 , + ∞) .

On this set we have the right to move from roots to powers:

x 1 9 · x · x 3 6 = x 1 9 · x · x 1 3 1 6

Using the properties of powers, we simplify the resulting power expression.

x 1 9 · x · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 3 1 6 = x 1 9 · x 1 6 · x 1 · 1 3 · 6 = = x 1 9 · x 1 6 x 1 18 = x 1 9 + 1 6 + 1 18 = x 1 3

Answer: x 1 9 · x · x 3 6 = x 1 3 .

Converting powers with variables in the exponent

These transformations are quite easy to make if you use the properties of the degree correctly. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.

We can replace by the product of powers, the exponents of which are the sum of some variable and a number. On the left side, this can be done with the first and last terms of the left side of the expression:

5 2 x 5 1 − 3 5 x 7 x − 14 7 2 x 7 − 1 = 0, 5 5 2 x − 3 5 x 7 x − 2 7 2 x = 0 .

Now let's divide both sides of the equality by 7 2 x. This expression for the variable x takes only positive values:

5 5 - 3 5 x 7 x - 2 7 2 x 7 2 x = 0 7 2 x , 5 5 2 x 7 2 x - 3 5 x 7 x 7 2 x - 2 7 2 x 7 2 x = 0 , 5 5 2 x 7 2 x - 3 5 x 7 x 7 x 7 x - 2 7 2 x 7 2 x = 0

Let's reduce fractions with powers, we get: 5 · 5 2 · x 7 2 · x - 3 · 5 x 7 x - 2 = 0.

Finally, the ratio of powers with the same exponents is replaced by powers of ratios, resulting in the equation 5 5 7 2 x - 3 5 7 x - 2 = 0, which is equivalent to 5 5 7 x 2 - 3 5 7 x - 2 = 0 .

Let us introduce a new variable t = 5 7 x, which reduces the solution of the original exponential equation to the solution quadratic equation 5 · t 2 − 3 · t − 2 = 0 .

Converting expressions with powers and logarithms

Expressions containing powers and logarithms are also found in problems. An example of such expressions is: 1 4 1 - 5 · log 2 3 or log 3 27 9 + 5 (1 - log 3 5) · log 5 3. The transformation of such expressions is carried out using the approaches and properties of logarithms discussed above, which we discussed in detail in the topic “Transformation of logarithmic expressions”.

If you notice an error in the text, please highlight it and press Ctrl+Enter

An algebraic expression in which, along with the operations of addition, subtraction and multiplication, also uses division into letter expressions, is called a fractional algebraic expression. These are, for example, the expressions

We call an algebraic fraction algebraic expression, which has the form of the quotient of the division of two integer algebraic expressions (for example, monomials or polynomials). These are, for example, the expressions

The third of the expressions).

Identical transformations of fractional algebraic expressions are for the most part intended to represent them in the form algebraic fraction. To find the common denominator, factorization of the denominators of fractions is used - terms in order to find their least common multiple. When reducing algebraic fractions, the strict identity of expressions may be violated: it is necessary to exclude values ​​of quantities at which the factor by which the reduction is made becomes zero.

Let us give examples of identical transformations of fractional algebraic expressions.

Example 1: Simplify an expression

All terms can be reduced to a common denominator (it is convenient to change the sign in the denominator of the last term and the sign in front of it):

Our expression is equal to one for all values ​​except these values; it is undefined and reducing the fraction is illegal).

Example 2. Represent the expression as an algebraic fraction

Solution. The expression can be taken as a common denominator. We find sequentially:

Exercises

1. Find the values ​​of algebraic expressions for the specified parameter values:

2. Factorize.

Simplifying algebraic expressions is one of the key points learning algebra and an extremely useful skill for all mathematicians. Simplification allows you to reduce a complex or long expression to a simple expression that is easy to work with. Basic skills of simplification are good even for those who are not enthusiastic about mathematics. By following a few simple rules, you can simplify many of the most common types of algebraic expressions without any special mathematical knowledge.

Steps

Important Definitions

1. Similar members. These are members with a variable of the same order, members with the same variables, or free members (members that do not contain a variable). In other words, similar terms include the same variable to the same degree, include several of the same variables, or do not include a variable at all. The order of the terms in the expression does not matter.

   • For example, 3x 2 and 4x 2 are similar terms because they contain a second-order (to the second power) variable "x". However, x and x2 are not similar terms, since they contain the variable “x” of different orders (first and second). Likewise, -3yx and 5xz are not similar terms because they contain different variables.

2. Factorization. This is finding numbers whose product leads to the original number. Any original number can have several factors. For example, the number 12 can be decomposed into next row factors: 1 × 12, 2 × 6 and 3 × 4, so we can say that the numbers 1, 2, 3, 4, 6 and 12 are factors of the number 12. Factors are the same as divisors, that is, the numbers by which the original number is divided .

   • For example, if you want to factor the number 20, write it like this: 4×5.
   • Note that when factoring, the variable is taken into account. For example, 20x = 4(5x).
   • Prime numbers cannot be factored because they are only divisible by themselves and 1.

3. Remember and follow the order of operations to avoid mistakes.

   • Brackets
   • Degree
   • Multiplication
   • Division
   • Addition
   • Subtraction

   Bringing similar members

   1. Write down the expression. Simple algebraic expressions (those that don't contain fractions, roots, etc.) can be solved (simplified) in just a few steps.

      • For example, simplify the expression 1 + 2x - 3 + 4x.

   2. Define similar terms (terms with a variable of the same order, terms with the same variables, or free terms).

      • Find similar terms in this expression. The terms 2x and 4x contain a variable of the same order (first). Also, 1 and -3 are free terms (do not contain a variable). Thus, in this expression the terms 2x and 4x are similar, and the members 1 and -3 are also similar.

   3. Give similar terms. This means adding or subtracting them and simplifying the expression.

      • 2x + 4x = 6x
      • 1 - 3 = -2

   4. Rewrite the expression taking into account the given terms. You will get a simple expression with fewer terms. The new expression is equal to the original one.

      • In our example: 1 + 2x - 3 + 4x = 6x - 2, that is, the original expression is simplified and easier to work with.

   5. Follow the order of operations when bringing similar members. In our example, it was easy to provide similar terms. However, in the case of complex expressions in which terms are enclosed in parentheses and fractions and roots are present, it is not so easy to bring such terms. In these cases, follow the order of operations.

      • For example, consider the expression 5(3x - 1) + x((2x)/(2)) + 8 - 3x. Here it would be a mistake to immediately define 3x and 2x as similar terms and present them, because it is necessary to open the parentheses first. Therefore, perform the operations according to their order.
        • 5(3x-1) + x((2x)/(2)) + 8 - 3x
        • 15x - 5 + x(x) + 8 - 3x
        • 15x - 5 + x 2 + 8 - 3x. Now, when the expression contains only addition and subtraction operations, you can bring similar terms.
        • x 2 + (15x - 3x) + (8 - 5)
        • x 2 + 12x + 3

   Taking the multiplier out of brackets

   1. Find the greatest common divisor (GCD) of all the coefficients of the expression. GCD is greatest number, by which all coefficients of the expression are divided.

      • For example, consider the equation 9x 2 + 27x - 3. In this case, GCD = 3, since any coefficient of this expression is divisible by 3.

   2. Divide each term of the expression by gcd. The resulting terms will contain smaller coefficients than in the original expression.

      • In our example, divide each term in the expression by 3.
        • 9x 2 /3 = 3x 2
        • 27x/3 = 9x
        • -3/3 = -1
        • The result was an expression 3x 2 + 9x - 1. It is not equal to the original expression.

   3. Write the original expression as equal to the product GCD of the resulting expression. That is, enclose the resulting expression in brackets, and take the gcd out of the brackets.

      • In our example: 9x 2 + 27x - 3 = 3(3x 2 + 9x - 1)

   4. Simplifying fractional expressions by putting the factor out of brackets. Why simply put the multiplier out of brackets, as was done earlier? Then, to learn how to simplify complex expressions, such as fractional expressions. In this case, putting the factor out of brackets can help get rid of the fraction (from the denominator).

      • For example, consider fractional expression(9x 2 + 27x - 3)/3. Use factoring out to simplify this expression.
        • Put the factor of 3 out of brackets (as you did earlier): (3(3x 2 + 9x - 1))/3
        • Notice that there is now a 3 in both the numerator and the denominator. This can be reduced to give the expression: (3x 2 + 9x – 1)/1
        • Since any fraction that has the number 1 in the denominator is simply equal to the numerator, the original fraction expression simplifies to: 3x 2 + 9x - 1.

   Additional simplification methods

4. Let's look at a simple example: √(90). The number 90 can be factored into the following factors: 9 and 10, and extracted from 9 Square root(3) and remove 3 from under the root.
   • √(90)
   • √(9×10)
   • √(9)×√(10)
   • 3×√(10)
   • 3√(10)

5. Simplifying expressions with powers. Some expressions contain operations of multiplication or division of terms with powers. In the case of multiplying terms with the same base, their powers are added; in the case of dividing terms with the same base, their powers are subtracted.

   • For example, consider the expression 6x 3 × 8x 4 + (x 17 /x 15). In the case of multiplication, add the powers, and in the case of division, subtract them.
     • 6x 3 × 8x 4 + (x 17 /x 15)
     • (6 × 8)x 3 + 4 + (x 17 - 15)
     • 48x 7 + x 2
   • The following is an explanation of the rules for multiplying and dividing terms with powers.
     • Multiplying terms with powers is equivalent to multiplying terms by themselves. For example, since x 3 = x × x × x and x 5 = x × x × x × x × x, then x 3 × x 5 = (x × x × x) × (x × x × x × x × x), or x 8 .
     • Likewise, dividing terms with degrees is equivalent to dividing terms by themselves. x 5 / x 3 = (x × x × x × x × x)/(x × x × x). Since similar terms found in both the numerator and the denominator can be reduced, the product of two “x”, or x 2 , remains in the numerator.

• Always remember about the signs (plus or minus) before the terms of the expression, as many people have difficulty choosing the correct sign.
• Ask for help if needed!
• Simplifying algebraic expressions isn't easy, but once you get the hang of it, it's a skill you can use for the rest of your life.

Section 5 EXPRESSIONS AND EQUATIONS

In this section you will learn:

ü o expressions and their simplifications;

ü what are the properties of equalities;

ü how to solve equations based on the properties of equalities;

ü what types of problems are solved using equations; what are perpendicular lines and how to build them;

ü what lines are called parallel and how to build them;

ü what is a coordinate plane?

ü how to determine the coordinates of a point on a plane;

ü what is a graph of the relationship between quantities and how to construct it;

ü how to apply the studied material in practice

§ 30. EXPRESSIONS AND THEIR SIMPLIFICATION

You already know what letter expressions are and know how to simplify them using the laws of addition and multiplication. For example, 2a ∙ (-4 b ) = -8 ab . In the resulting expression, the number -8 is called the coefficient of the expression.

Does the expression CD coefficient? So. It is equal to 1 because cd - 1 ∙ cd .

Recall that converting an expression with parentheses into an expression without parentheses is called expanding the parentheses. For example: 5(2x + 4) = 10x+ 20.

The reverse action in this example is to take the common factor out of brackets.

Terms containing the same letter factors are called similar terms. By taking the common factor out of brackets, similar terms are raised:

5x + y + 4 - 2x + 6 y - 9 =

= (5x - 2x) + (y + 6 y )+ (4 - 9) = = (5-2)* + (1 + 6)* y -5 =

B x+ 7y - 5.

Rules for opening parentheses

1. If there is a “+” sign in front of the brackets, then when opening the brackets, the signs of the terms in the brackets are preserved;

2. If there is a “-” sign in front of the brackets, then when the brackets are opened, the signs of the terms in the brackets change to the opposite.

Task 1. Simplify the expression:

1) 4x+(-7x + 5);

2) 15 y -(-8 + 7 y ).

Solutions. 1. Before the brackets there is a “+” sign, so when opening the brackets, the signs of all terms are preserved:

4x +(-7x + 5) = 4x - 7x + 5=-3x + 5.

2. Before the brackets there is a “-” sign, so when opening the brackets: the signs of all terms are reversed:

15 - (- 8 + 7y) = 15y + 8 - 7y = 8y +8.

To open the parentheses, use the distributive property of multiplication: a( b + c ) = ab + ac. If a > 0, then the signs of the terms b and with do not change. If a< 0, то знаки слагаемых b and change to the opposite.

Task 2. Simplify the expression:

1) 2(6 y -8) + 7 y ;

2)-5(2-5x) + 12.

Solutions. 1. The factor 2 in front of the brackets is positive, therefore, when opening the brackets, we preserve the signs of all terms: 2(6 y - 8) + 7 y = 12 y - 16 + 7 y =19 y -16.

2. The factor -5 in front of the brackets is negative, so when opening the brackets, we change the signs of all terms to the opposite:

5(2 - 5x) + 12 = -10 + 25x +12 = 2 + 25x.

Find out more

1. The word “sum” comes from Latin summa , which means “total”, “total amount”.

2. The word “plus” comes from Latin plus which means "more" and the word "minus" is from Latin minus What does "less" mean? The signs “+” and “-” are used to indicate the operations of addition and subtraction. These signs were introduced by the Czech scientist J. Widman in 1489 in the book “A quick and pleasant account for all merchants”(Fig. 138).

Rice. 138

REMEMBER THE IMPORTANT

1. What terms are called similar? How are similar terms constructed?

2. How do you open parentheses preceded by a “+” sign?

3. How do you open parentheses preceded by a “-” sign?

4. How do you open parentheses preceded by a positive factor?

5. How do you open parentheses that are preceded by a negative factor?

1374". Name the coefficient of the expression:

1)12 a; 3) -5.6 xy;

2)4 6; 4)-s.

1375". Name the terms that differ only by coefficient:

1) 10a + 76-26 + a; 3) 5 n + 5 m -4 n + 4;

2) bc -4 d - bc + 4 d ; 4)5x + 4y-x + y.

What are these terms called?

1376". Are there any similar terms in the expression:

1)11a+10a; 3)6 n + 15 n ; 5) 25r - 10r + 15r;

2) 14s-12; 4)12 m + m ; 6)8 k +10 k - n ?

1377". Is it necessary to change the signs of the terms in brackets, opening the brackets in the expression:

1)4 + (a+ 3 b); 2)-c +(5-d); 3) 16-(5 m -8 n)?

1378°. Simplify the expression and underline the coefficient:

1379°. Simplify the expression and underline the coefficient:

1380°. Combine similar terms:

1) 4a - Po + 6a - 2a; 4) 10 - 4 d - 12 + 4 d ;

2) 4 b - 5 b + 4 + 5 b ; 5) 5a - 12 b - 7a + 5 b;

3)-7 ang="EN-US">c+ 5-3 c + 2; 6) 14 n - 12 m -4 n -3 m.

1381°. Combine similar terms:

1) 6a - 5a + 8a -7a; 3) 5s + 4-2s-3s;

2)9 b +12-8-46; 4) -7 n + 8 m - 13 n - 3 m.

1382°. Take the common factor out of brackets:

1)1.2 a +1.2 b; 3) -3 n - 1.8 m; 5)-5 p + 2.5 k -0.5 t ;

2) 0.5 s + 5 d; 4) 1.2 n - 1.8 m; 6) -8r - 10k - 6t.

1383°. Take the common factor out of brackets:

1) 6a-12 b; 3) -1.8 n -3.6 m;

2) -0.2 s + 1 4 d ; A) 3p - 0.9 k + 2.7 t.

1384°. Open the brackets and combine similar terms;

1) 5 + (4a -4); 4) -(5 c - d) + (4 d + 5c);

2) 17x-(4x-5); 5) (n - m) - (-2 m - 3 n);

3) (76 - 4) - (46 + 2); 6) 7(-5x + y) - (-2y + 4x) + (x - 3y).

1385°. Open the brackets and combine similar terms:

1) 10a + (4 - 4a); 3) (s - 5 d) - (- d + 5c);

2) -(46- 10) + (4- 56); 4)-(5 n + m) + (-4 n + 8 m)-(2 m -5 n).

1386°. Open the brackets and find the meaning of the expression:

1)15+(-12+ 4,5); 3) (14,2-5)-(12,2-5);

2) 23-(5,3-4,7); 4) (-2,8 + 13)-(-5,6 + 2,8) + (2,8-13).

1387°. Open the brackets and find the meaning of the expression:

1) (14- 15,8)- (5,8 + 4);

2)-(18+22,2)+ (-12+ 22,2)-(5- 12).

1388°. Open parenthesis:

1)0.5 ∙ (a + 4); 4) (n - m) ∙ (-2.4 p);

2)-s ∙ (2.7-1.2 d ); 5)3 ∙ (-1.5 r + k - 0.2 t);

3) 1.6 ∙ (2 n + m); 6) (4.2 p - 3.5 k -6 t) ∙ (-2a).

1389°. Open parenthesis:

1) 2.2 ∙ (x-4); 3)(4 c - d )∙(-0.5 y );

2) -2 ∙ (1.2 n - m); 4)6- (-р + 0.3 k - 1.2 t).

1390. Simplify the expression:

1391. Simplify the expression:

1392. Reduce similar terms:

1393. Combine similar terms:

1394. Simplify the expression:

1)2.8 - (0.5 a + 4) - 2.5 ∙ (2a - 6);

2) -12 ∙ (8 - 2, by ) + 4.5 ∙ (-6 y - 3.2);

4) (-12.8 m + 24.8 n) ∙ (-0.5)-(3.5 m -4.05 m) ∙ 2.

1395. Simplify the expression:

1396. Find the meaning of the expression;

1) 4-(0.2 a-3)-(5.8 a-16), if a = -5;

2) 2-(7-56)+ 156-3∙(26+ 5), if = -0.8;

m = 0.25, n = 5.7.

1397. Find the meaning of the expression:

1) -4∙ (i-2) + 2∙(6x - 1), if x = -0.25;

1398*. Find the error in the solution:

1)5- (a-2.4)-7 ∙ (-a+ 1.2) = 5a - 12-7a + 8.4 = -2a-3.6;

2) -4 ∙ (2.3 a - 6) + 4.2 ∙ (-6 - 3.5 a) = -9.2 a + 46 + 4.26 - 14.7 a = -5.5 a + 8.26.

1399*. Open the parentheses and simplify the expression:

1) 2ab - 3(6(4a - 1) - 6(6 - 10a)) + 76;

1400*. Arrange the parentheses to get the correct equality:

1)a-6-a + 6 = 2a; 2) a -2 b -2 a + b = 3 a -3 b .

1401*. Prove that for any numbers a and b if a > b , then the equality holds:

1) (a + b) + (a- b) = 2a; 2) (a + b) - (a - b) = 2 b.

Will this equality be correct if: a) a< b ; b) a = 6?

1402*. Prove that for any natural number and the arithmetic mean of the previous and following numbers is equal to the number a.

PUT IT IN PRACTICE

1403. To prepare a fruit dessert for three people you need: 2 apples, 1 orange, 2 bananas and 1 kiwi. How to create a letter expression to determine the amount of fruit needed to prepare dessert for guests? Help Marin calculate how many fruits she needs to buy if: 1) 5 friends come to visit her; 2) 8 friends.

1404. Make a letter expression to determine the time required to complete your math homework if:

1) a min was spent on solving problems; 2) simplification of expressions is 2 times greater than for solving problems. How long did it take to complete homework Vasilko, if he spent 15 minutes solving problems?

1405. Lunch in the school cafeteria consists of salad, borscht, cabbage rolls and compote. The cost of salad is 20%, borscht - 30%, cabbage rolls - 45%, compote - 5% total cost just lunch. Write an expression to find the cost of lunch in the school canteen. How much does lunch cost if the price of salad is 2 UAH?

REVIEW PROBLEMS

1406. Solve the equation:

1407. Tanya spent on ice creamall available money, and for candy -the rest. How much money does Tanya have left?

if candy costs 12 UAH?