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

That is, if you substitute some (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 decomposed into factors).

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

To fix it yourself, a few examples:

Examples:

Solutions:

1. I hope you did not immediately rush to cut and? It was still not enough to “reduce” units like this:

The first step should be to factorize:

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

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

Let's remember:

Answers:

1. The denominators and are coprime, that is, they do not have 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 mixed fractions into improper 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 numerical fractions: we find a 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 factor them:

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 we write out all the common factors once;

and multiply them by all other factors, not common ones.

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

We emphasize the common factors:

Now we write out the common factors once and add to them all non-common (not underlined) factors:

This is the common denominator.

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

We decompose the denominators into factors;

determine common (identical) multipliers;

write out all the common factors once;

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

So, in order:

1) decompose the denominators into factors:

2) determine the common (identical) factors:

3) write out all the common factors once and multiply them by all the other (not underlined) 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 example: .

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 to make fractions have 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 it's 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 you bring fractions to a common denominator, use only the multiplication operation!

But what do you need to multiply to get?

Here on and multiply. And multiply by:

Expressions that cannot be factorized will be called "elementary factors".

For example, is an elementary factor. - Same. But - no: it is decomposed into factors.

What about expression? Is it elementary?

No, because it can be factorized:

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

So, the elementary factors into which you decompose an expression with letters are an analogue of the simple factors into which you decompose numbers. And we will do the same with them.

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

The multiplier is elementary, and they do not have it in common, 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? Both of them represent:

Great! Then:

Another example:

Solution:

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

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

So let's write:

That is, it turned out like this: inside the bracket, 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? Now let's check.

Tasks for independent solution:

Answers:

Here we must remember one more thing - the difference of cubes:

Please note that the denominator of the second fraction does not contain the formula "square of the sum"! 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 last, and not their doubled product. The incomplete square of the sum is one of the factors in the expansion of the difference of cubes:

What if there are already three fractions?

Yes, the same! First of all, we will make sure that the maximum number of factors in the denominators is the same:

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

We write out the first denominator in full in the common denominator, and then we 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 goes like this:

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

It's simple: you know how to add fractions, right? So, you need to make sure that the deuce becomes 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 will turn into a fraction:

Exactly what is needed!

5. Multiplication and division of fractions.

Well, the hardest part is now over. 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, considering the value of such an expression:

Did you count?

It should work.

So, I 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 perform addition and subtraction. Again, in any order.

But: the parenthesized expression is evaluated out of order!

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

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

So, the order of actions 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 that's not the same as an expression with letters, is it?

No, it's the same! Only instead of arithmetic operations it is necessary to do algebraic operations, that is, the operations described in the previous section: bringing similar, adding fractions, reducing fractions, and so on. The only difference will be the action of factoring polynomials (we often use it when working with fractions). Most often, for factorization, you need to use i or simply take the common factor out of brackets.

Usually our goal is to represent an expression as a product or quotient.

For example:

Let's simplify the expression.

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

It is impossible to simplify this expression further, all 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 look at the solution.

Solution:

First of all, let's define the procedure.

First, let's add the fractions in brackets, instead of two fractions, one will turn out.

Then we will do the division of fractions. Well, we add the result with the last fraction.

I will schematically number the steps:

Now I will show the whole process, tinting the current action with 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 goes for reducing fractions: as soon as an opportunity arises 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 (brief):

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

Now on to learning!

EXPRESSION CONVERSION. SUMMARY AND BASIC FORMULA

Basic simplification operations:

• Bringing similar: to add (reduce) like terms, you need to add their coefficients and assign the letter part.
• Factorization: taking the common factor out of brackets, applying, etc.
• Fraction reduction: the numerator and denominator of a fraction can be multiplied or divided by the same non-zero number, from which the value of the fraction does not change.
  1) numerator and denominator factorize
  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:
  ;

An expression of the form a (m/n) , where n is some natural number, m is some integer and the base of degree a is greater than zero, is called a degree with a fractional exponent. Moreover, the following equality is true. n√(a m) = a (m/n) .

As we already know, numbers of the form m/n, where n is some natural number and m is some integer, are called fractional or rational numbers. From the above, we get that the degree is defined, for any rational exponent and any positive base of the degree.

For any rational numbers p,q and any a>0 and b>0, the following equalities are true:

• 1. (a p)*(a q) = a (p+q)
• 2. (a p): (b q) = a (p-q)
• 3. (a p) q = a (p*q)
• 4. (a*b) p = (a p)*(b p)
• 5. (a/b) p = (a p)/(b p)

These properties are widely used when converting various expressions that contain degrees with fractional exponents.

Examples of transformations of expressions containing a degree with a fractional exponent

Let's look at a few examples that demonstrate how these properties can be used to transform expressions.

1. Calculate 7 (1/4) * 7 (3/4) .

• 7 (1/4) * 7 (3/4) = z (1/4 + 3/4) = 7.

2. Calculate 9 (2/3) : 9 (1/6) .

• 9 (2/3) : 9 (1/6) = 9 (2/3 - 1/6) = 9 (1/2) = √9 = 3.

3. Calculate (16 (1/3)) (9/4) .

• (16 (1/3)) (9/4) = 16 ((1/3)*(9/4)) =16 (3/4) = (2 4) (3/4) = 2 (4*3/4) = 2 3 = 8.

4. Calculate 24 (2/3) .

• 24 (2/3) = ((2 3)*3) (2/3) = (2 (2*2/3))*3 (2/3) = 4*3√(3 2)=4*3√9.

5. Calculate (8/27) (1/3) .

• (8/27) (1/3) = (8 (1/3))/(27 (1/3)) = ((2 3) (1/3))/((3 3) (1/3))= 2/3.

6. Simplify the expression ((a (4/3))*b + a*b (4/3))/(3√a + 3√b)

• ((a (4/3))*b + a*b (4/3))/(3√a + 3√b) = (a*b*(a (1/3) + b (1/3 )))/(1/3) + b (1/3)) = a*b.

7. Calculate (25 (1/5))*(125 (1/5)).

• (25 (1/5))*(125 (1/5)) =(25*125) (1/5) = (5 5) (1/5) = 5.

8. Simplify the expression

• (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a(2/3) + a(-1/3)).
• (a (1/3) - a (7/3))/(a (1/3) - a (4/3)) - (a (-1/3) - a (5/3))/( a(2/3) + a(-1/3)) =
• = ((a (1/3))*(1-a 2))/((a (1/3))*(1-a)) - ((a (-1/3))*(1- a 2))/ ((a (-1/3))*(1+a)) =
• = 1 + a - (1-a) = 2*a.

As you can see, using these properties, you can greatly simplify some expressions that contain degrees with fractional exponents.

Let's consider the topic of transforming expressions with powers, but first we will dwell on a number of transformations that can be carried out with any expressions, including power ones. We will learn how to open brackets, give like terms, work with the base and exponent, use the properties of powers.

What are Power Expressions?

In the school course, few people use the phrase "power expressions", but this term is constantly found in collections for preparing for the exam. In most cases, the 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 degrees.

We give several examples of power expressions, starting with a degree with a natural exponent and ending with a degree 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 . As well as 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 a variable 3 x - 54 - 7 3 x - 58 or a 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 take a look at their transformation.

The main types of transformations of power expressions

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

Example 1

Calculate Power Expression Value 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 between the two numbers. We have 2 3 (4 2 − 12) = 2 3 (16 − 12) = 2 3 4.

It remains for us to replace the degree 2 3 its meaning 8 and calculate the product 8 4 = 32. Here is our answer.

Answer: 2 3 (4 2 − 12) = 32 .

Example 2

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

Solution

The expression given to us in the condition of the problem contains similar terms, which we can bring: 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 an expression with powers of 9 - b 3 · π - 1 2 as a product.

Solution

Let's represent 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 .

And now let's move on to the analysis of identical 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 . It is difficult to work with such records. 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.

The transformations of the degree and the indicator are carried out according to the rules known to us separately from each other. The most important thing is that as a result of the transformations, an expression is obtained that is identical to the original one.

The purpose of transformations is to simplify the original expression or to obtain a solution to the problem. For example, in the example we gave above, (2 + 0 , 3 7) 5 − 3 , 7 you can perform operations to go to the degree 4 , 1 1 , 3 . Opening the brackets, we can bring like terms in the base of the degree (a (a + 1) − a 2) 2 (x + 1) and get a power expression of a simpler form a 2 (x + 1).

Using Power Properties

The properties of degrees, written as equalities, are one of the main tools for transforming expressions with degrees. We present here the main ones, considering 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 stringent. 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.

You can apply the properties of degrees without restrictions in cases where the bases of the degrees are positive or contain variables whose range of acceptable values ​​is such that the bases take only positive values ​​on it. In fact, within the framework of the school curriculum in mathematics, the task of the student is to choose the appropriate property and apply it correctly.

When preparing for admission to universities, there may be tasks in which inaccurate application of properties will lead to a narrowing of the ODZ and other difficulties with the solution. In this section, we will consider only two such cases. More information on the subject can be found in the topic "Transforming expressions using exponent properties".

Example 4

Represent the expression a 2 , 5 (a 2) - 3: a - 5 , 5 as a degree with a base a.

Solution

To begin with, we use the exponentiation property 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 .

The transformation of power expressions according to the property of degrees 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 the equality (a b) r = a r b r, from right to left, then 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 \u003d 21 1 3 + 2 3 \u003d 21 1 \u003d 21.

There is another way to make transformations:

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

Imagine the degree a 1 , 5 How a 0 , 5 3. Using the degree property in a degree (a r) s = a r s from right to left and get (a 0 , 5) 3: a 1 , 5 - a 0 , 5 - 6 = (a 0 , 5) 3 - a 0 , 5 - 6 . In the resulting expression, you can easily enter a new variable t = a 0 , 5: get t 3 − t − 6.

Answer: t 3 − t − 6 .

Converting fractions containing powers

We usually deal with two variants of power expressions with fractions: the expression is a fraction with a degree or contains such a fraction. All basic fraction transformations are applicable to such expressions without restrictions. They can be reduced, brought to a new denominator, work 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

Put a minus 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 vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example 8

Bring 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) We choose 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 take a 0 , 3. The range of admissible values ​​of the variable a includes the set of all positive real numbers. In this area, the degree 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) 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

Multiply this expression by x 1 3 + 2 · y 1 6 , we get the sum of 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 bring the original fraction.

So we found an additional factor x 1 3 + 2 · y 1 6 . On the range of acceptable values ​​of variables x And y the expression x 1 3 + 2 y 1 6 does not vanish, so 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) Use the greatest common denominator (GCD) by which the numerator and denominator can be reduced. For the numbers 30 and 45, this is 15 . We can also reduce x 0 , 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 .

The basic operations with fractions include reduction to a new denominator and reduction of fractions. Both actions are performed in compliance with a number of rules. When adding and subtracting fractions, the fractions are first reduced to a common denominator, after which actions (addition or subtraction) are performed with 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 brackets. 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 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 degree 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 formula for the difference of squares: 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 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 a fraction x 3 4 x - 5 8 x 2, 7 + 1.

Let's continue transformations of x powers x 3 4 x - 5 8 · 1 x 2 , 7 + 1 . Now you can use the power division property 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 pass 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 multipliers with negative exponents from the numerator to the denominator and vice versa by changing the sign of the exponent. This action simplifies 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 tasks, there are power expressions that contain not only degrees with fractional exponents, but also roots. It is desirable to reduce such expressions only to roots or only to powers. The transition to degrees is preferable, since they are easier to work with. Such a transition is especially advantageous when the DPV of the variables for the original expression allows you to replace the roots with powers without having to access the modulus or split the DPV into several intervals.

Example 12

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

Solution

Valid range of a variable x is determined 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 degrees, 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 simple to make if you correctly use the properties of the degree. For example, 5 2 x + 1 − 3 5 x 7 x − 14 7 2 x − 1 = 0.

We can replace the product of the degree, in terms of which the sum of some variable and a number is found. On the left side, this can be done with the first and last terms on 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 equation by 7 2 x. This expression on the ODZ of 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 the 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, which leads to 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 .

We introduce a new variable t = 5 7 x , which reduces the solution of the original exponential equation to the solution of the 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. Examples of such expressions are: 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 discussed above and the properties of logarithms, which we have analyzed in detail in the topic “Transformation of logarithmic expressions”.

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Expressions, expression conversion

Power expressions (expressions with powers) and their transformation

In this article, we will talk about transforming expressions with powers. First, we will focus on transformations that are performed with expressions of any kind, including power expressions, such as opening brackets, reducing 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 textbooks of mathematics, but it often appears in collections of problems, especially designed to prepare for the Unified State Examination and the OGE, for example,. After analyzing tasks in which it is required to perform any actions with power expressions, it becomes clear that power expressions are understood as expressions containing degrees in their entries. Therefore, for yourself, you can take the following definition:

Definition.

Power expressions are expressions containing powers.

Let's bring examples of power expressions. Moreover, we will represent them according to how the development of views on from a degree with a natural indicator to a degree with a real indicator takes place.

As you know, first you get acquainted with the degree 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 −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 negative integer powers, like the following: 3 −2, , a −2 +2 b −3 + c 2 .

In the senior classes, they return to the degrees again. There, a degree with a rational exponent is introduced, which leads to 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 there are, for example, such expressions 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 figured out the question of what are power expressions. Next, we will learn how to transform them.

The main types of transformations of power expressions

With power expressions, you can perform any of the basic identical transformations of expressions. For example, you can open brackets, replace numeric expressions with their values, add like terms, and so on. Naturally, in this case it is necessary to comply with the accepted order of actions. Let's give examples.

Example.

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

Solution.

According to the order of the actions, we first perform the actions in brackets. There, firstly, we replace the power of 4 2 with its value 16 (see if necessary), 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 of 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 Power Expressions 3 a 4 b −7 −1+2 a 4 b −7.

Solution.

Obviously, this expression contains like terms 3 a 4 b −7 and 2 a 4 b −7 , and we can reduce 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.

To cope with the task allows the representation of the number 9 as a power of 3 2 and subsequent use abbreviated multiplication formulas difference of squares:

Answer:

There are also a number of identical transformations inherent in power expressions. Next, we will analyze them.

Working with base and exponent

There are degrees, in the basis and / or indicator of which are not just numbers or variables, but some expressions. As an example, let's write (2+0.3 7) 5−3.7 and (a (a+1)−a 2) 2 (x+1) .

When working with similar expressions, both the expression in the base of the degree and the expression in the exponent can be replaced by an identically equal expression on ODZ his variables. In other words, according to the rules known to us, we can separately convert the base of the degree, and separately - the indicator. It is clear that as a result of this transformation, an expression is 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 (2+0.3 7) 5−3.7 mentioned above, you can perform operations with numbers in the base and exponent, which will allow you to go to the power of 4.1 1.3. And after opening the brackets and bringing like terms in the base of the degree (a·(a+1)−a 2) 2·(x+1) we get a power expression of a simpler form a 2·(x+1) .

Using Power 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 power properties hold:

• 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, 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 ones, and for a=0 .

At school, the main attention in the transformation of power expressions is focused precisely on the ability to choose the appropriate property and apply it correctly. In this case, the bases of the degrees are usually positive, which allows you to use the properties of the degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the range of acceptable values ​​​​of variables is usually such that the bases take only positive values ​​on it, which allows you to freely use the properties of degrees. In general, you need to constantly ask yourself whether it is possible to apply any property of degrees in this case, because inaccurate use of properties can lead to a narrowing of the ODZ and other troubles. These points are discussed in detail and with examples in the article. transformation of expressions using the properties of degrees. Here we confine ourselves to 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 by the property of raising a power to a power: (a 2) −3 =a 2 (−3) =a −6. In this case, the initial 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 \u003d a 2.

Power properties are used when transforming power expressions both from left to right and from right to left.

Example.

Find the value of the power expression.

Solution.

Equality (a·b) r =a r ·b r , applied from right to left, allows you to go from the original expression to the product of the form and further. And when multiplying powers with the same base, the indicators add up: .

It was possible to perform the transformation of the original expression in another way:

Answer:

.

Example.

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

Solution.

The degree a 1.5 can be represented as a 0.5 3 and further on the basis of the property of the degree in the degree (a r) s =a r s applied from right to left, convert 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 is 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 fractions with powers or represent such fractions. To such fractions, any of the main fraction conversions, which are inherent in fractions of any kind. That is, fractions that contain degrees can be reduced, reduced to a new denominator, work separately with their numerator and separately with the denominator, etc. To illustrate the above words, consider the solutions of 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 expression obtained after that using the properties of powers, and in the denominator we present similar terms:

And we 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. At the same time, 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 DPV. To prevent this from happening, it is necessary that the additional factor does not vanish for any values ​​of the variables from the ODZ variables for the original expression.

Example.

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

Solution.

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

b) Looking more closely at the denominator, we 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 bring the original fraction.

So we found an additional factor . The expression does not vanish on the range of acceptable values ​​of the variables x and y, 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 degrees: the numerator and denominator are represented as a certain number of factors, and the same factors of the numerator and denominator are reduced.

Example.

Reduce the fraction: a) , b).

Solution.

a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which equals 15. Also, obviously, you can reduce by x 0.5 +1 and by . Here's what we have:

b) In this case, the same factors in the numerator and denominator are not immediately visible. To get them, you have to perform preliminary transformations. In this case, they consist in decomposing the denominator into factors according to the difference of squares formula:

Answer:

A)

b) .

Reducing fractions to a new denominator and reducing fractions is mainly used to perform operations on 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), and 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 reciprocal.

Example.

Follow the steps .

Solution.

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

Now we multiply fractions:

Obviously, a reduction by the power x 1/2 is possible, after which we have .

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

Answer:

Example.

Simplify 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 convert the resulting fraction into a product. This gives us the opportunity to use the property of dividing powers with the same bases: . And at the end of the process, we pass from the last product to the fraction.

Answer:

.

And we 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 by 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, along with degrees with fractional exponents, there are also roots. To convert such an expression to the desired form, in most cases it is enough to go only to roots or only to powers. But since it is more convenient to work with degrees, they usually move from roots to degrees. 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 degrees without the need to access the module or split the ODZ into several intervals (we discussed this in detail in the article, the transition from roots to powers and vice versa After getting acquainted with the degree with a rational exponent a degree with an irrational indicator is introduced, which makes it possible to speak of a degree with an arbitrary real indicator.At this stage, the school begins to study exponential function, which is analytically given by the degree, in the basis of which there is a number, and in the indicator - a variable. So we are faced with power expressions containing numbers in the base of the degree, 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 transformations are quite simple. In the vast majority of cases, they are based on the properties of the degree and are aimed mostly 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.

First, the exponents, in whose exponents the sum of some variable (or expression with variables) and a number, is found, 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 takes only positive values ​​on the ODZ variable x for the original equation (this is a standard technique for solving equations of this kind, we are not talking about it now, so focus on subsequent transformations of expressions with powers ):

Now fractions with powers are cancelled, which gives .

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

I. V. Boikov, L. D. Romanova Collection of tasks for preparing for the exam. Part 1. Penza 2003.