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

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

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

To fix the solution yourself, take a few examples:

Examples:

Solutions:

1. I hope you didn't rush to cut u right away? It was still not enough to "cut" units like this:

The first action should be factoring:

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

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

Let's remember:

Answers:

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

2. Here the common denominator is:

3. First thing here mixed fractions we turn them into wrong ones, and then - according to the usual scheme:

It is completely different if the fractions contain letters, for example:

Let's start simple:

a) Denominators do not contain letters

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

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

Try it yourself:

Answers:

b) Denominators contain letters

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

· First of all, we determine the common factors;

· Then write out all common factors one time;

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

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

Let's emphasize the common factors:

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

This is the common denominator.

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

· We decompose the denominators into factors;

· Determine common (identical) factors;

· Write out all common factors once;

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

So, in order:

1) we decompose the denominators into factors:

2) we determine the common (identical) factors:

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

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

By the way, there is one trick:

For 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 do you make fractions the same denominator?

Let's remember the basic property of a fraction:

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

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

So, another unshakable rule:

When you bring fractions to common denominator, use only the multiplication operation!

But what must be multiplied by in order to receive?

Here on and multiply. And multiply by:

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

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

What do you think about expression? Is it elementary?

No, since it can be factorized:

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

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

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

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

Another example:

Solution:

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

Excellent! Then:

Another example:

Solution:

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

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

So we will write:

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

Now we bring to a common denominator:

Got it? Let's check it out now.

Tasks for independent solution:

Answers:

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

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

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

What if there are already three fractions?

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

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

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

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

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

Exactly what is needed!

5. Multiplication and division of fractions.

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

Procedure

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

Have you counted?

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, they can be done in any order.

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

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

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

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

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

Okay, it's all simple.

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

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

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

For example:

Let's simplify the expression.

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

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

2) We get:

Multiplication of fractions: what could be easier.

3) Now you can shorten:

That's it. Nothing complicated, right?

Another example:

Simplify the expression.

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

Solution:

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

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

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

I will schematically number the actions:

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

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

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

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

And promised at the very beginning:

Answers:

Solutions (concise):

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

Now forward to learning!

TRANSFORMATION OF EXPRESSIONS. SUMMARY AND BASIC FORMULAS

Basic simplification operations:

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

  IMPORTANT: only multipliers can be reduced!

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

Topic: " Converting expressions containing exponents with fractional exponents "

“Let someone try to erase degrees from mathematics, and he will see that without them you cannot go far.” (M.V. Lomonosov)

Lesson objectives:

educational: to generalize and systematize the knowledge of students on the topic "Degree with a rational indicator", to monitor the level of assimilation of the material, to eliminate gaps in the knowledge and skills of students;

developing: form students' self-control skills; create an atmosphere of interest of each student in work, develop cognitive activity students;

educational: foster interest in the subject, in the history of mathematics.

Lesson type: lesson in generalization and systematization of knowledge

Equipment: grade sheets, cards with assignments, decoders, crosswords for each student.

Preliminary preparation: the class is divided into groups, in each group the leader is a consultant.

DURING THE CLASSES

I. Organizing time.

Teacher: We have completed the study of the topic "Degree with a rational exponent and its properties." Your task in this lesson is to show how you learned the studied material and how you are able to apply the knowledge gained when solving specific tasks... Each of you has a score sheet on the table. In it you will enter your grade for each stage of the lesson. At the end of the lesson, you will expose average score per lesson.

Evaluation paper

	Crossword	Warm up	Work in notebooks	Equations	Check yourself (s \ r)

II. Examination homework.

Mutual examination with a pencil in hand, the answers are read out by the students.

III. Updating students' knowledge.

Teacher: The famous French writer Anatole France said at one time: "Learning must be fun. ... To absorb knowledge, one must absorb it with appetite."

Let's repeat the necessary theoretical information in the course of solving the crossword puzzle.

Horizontally:

1. The action by which the degree value is calculated (erection).

2. A product consisting of the same factors (degree).

3. The effect of exponents when raising a degree to a degree (work).

4. Action of degrees at which exponents are subtracted (division).

Vertically:

5. The number of all the same factors (index).

6. Degree with zero exponent (unit).

7. Duplicate multiplier (base).

8. Value 10 5: (2 3 5 5) (four).

9. An exponent that is not usually written (unit).

IV. Mathematical warm-up.

Teacher. Let's repeat the definition of the degree with a rational exponent and its properties, we will perform the following tasks.

1. Represent the expression x 22 as a product of two degrees with base x, if one of the factors is: x 2, x 5.5, x 1 \ 3, x 17.5, x 0

2. Simplify:

b) y 5 \ 8 y 1 \ 4: y 1 \ 8 = y

c) s 1.4 s -0.3 s 2.9

3. Calculate and form a word using a decoder.

After completing this task, you guys will learn the name of the German mathematician who introduced the term “exponent”.

1) (-8) 1\3 2) 81 1\2 3) (3\5) -1 4) (5\7) 0 5) 27 -1\3 6) (2\3) -2 7) 16 1\2 * 125 1\3

Word: 1234567 (Pin)

V. Paperwork in notebooks (answers open on the board) .

Tasks:

1. Simplify the expression:

(x-2): (x 1 \ 2 -2 1 \ 2) (y-3): (y 1 \ 2 - 3 1 \ 2) (x-1): (x 2 \ 3 -x 1 \ 3 +1)

2. Find the value of the expression:

(x 3 \ 8 x 1 \ 4 :) 4 at x = 81

Vi. Group work.

The task. Solve equations and form a word using a decoder.

Card number 1

Word: 1234567 (Diophantus)

Card number 2

Card number 3

Word: 123451 (Newton)

Decoder

Teacher. All these scholars have contributed to the development of the concept of "degree".

Vii. Historical background on the development of the concept of degree (student message).

The concept of a degree with a natural indicator was formed even among the ancient peoples. Square and cube numbers were used to calculate areas and volumes. The degrees of some numbers were used by scientists to solve certain problems. Ancient egypt and Babylon.

In the III century, the book of the Greek scientist Diophantus "Arithmetic" was published, which laid the foundation for the introduction of alphabetic symbolism. Diophantus introduces symbols for the first six powers of the unknown and their reciprocal values. In this book, a square is denoted by a sign with an index r; the cube is by the sign k with the index r, and so on.

From the practice of solving more complex algebraic problems and operating with degrees, it became necessary to generalize the concept of a degree and expand it by introducing zero, negative and fractional numbers as an exponent. The idea of ​​generalizing the concept of a degree to a degree with an unnatural exponent of mathematics came gradually.

Fractional exponents and most simple rules actions on powers with fractional exponents are found in the French mathematician Nicholas Orem (1323–1382) in his work "Algorithm of Proportions".

Equality, and 0 = 1 (for and not equal to 0) was used in his writings at the beginning of the 15th century by the Samarkand scientist Giyasaddin Kashi Dzhemshid. Independently of him, the zero indicator was introduced by Nikolai Shuke in the 15th century. It is known that Nikolai Shuke (1445–1500) considered degrees with negative and zero exponents.

Later, fractional and negative exponents are found in “Complete arithmetic” (1544) by the German mathematician M. Stiefel and in Simon Stevin. Simon Stevin suggested to mean a 1 / n root.

The German mathematician M. Stiefel (1487–1567) defined a 0 = 1 at and introduced the name of the exponent (this is a literal translation from the German Exponent). German potenzieren means exponentiation.

At the end of the sixteenth century, François Viet introduced letters to denote not only variables, but also their coefficients. He used abbreviations: N, Q, C - for the first, second and third degrees. But modern designations (such as a 4, a 5) in the XVII were introduced by Rene Descartes.

Modern definitions and notation of degrees with zero, negative and fractional exponents originate from the works of the English mathematicians John Wallis (1616-1703) and Isaac Newton (1643-1727).

The expediency of introducing zero, negative and fractional exponents and modern symbols was first written in detail in 1665 by the English mathematician John Wallis. His business was completed by Isaac Newton, who began to systematically apply new symbols, after which they entered general use.

The introduction of a degree with a rational exponent is one of many examples of generalizing the concepts of mathematical action. A degree with zero, negative and fractional exponents is determined in such a way that the same rules of action are applied to it that take place for a degree with a natural exponent, i.e. so that the basic properties of the original definite concept of degree are preserved.

The new definition of a degree with a rational exponent does not contradict the old definition of a degree with a natural exponent, that is, the meaning of a new definition of a degree with a rational exponent is preserved for the particular case of a degree with a natural exponent. This principle, observed when generalizing mathematical concepts, is called the principle of permanence (preservation of constancy). It was expressed in an imperfect form in 1830 by the English mathematician J. Peacock; it was fully and clearly established by the German mathematician G. Hankel in 1867.

VIII. Check yourself.

Independent work by cards (answers open on the board) .

Option 1

1. Calculate: (1 point)

(a + 3a 1 \ 2): (a 1 \ 2 +3)

Option 2

1. Calculate: (1 point)

2. Simplify the expression: 1 point each

a) x 1.6 x 0.4 b) (x 3 \ 8) -5 \ 6

3. Solve the equation: (2 points)

4. Simplify the expression: (2 points)

5. Find the value of the expression: (3 points)

IX. Summing up the lesson.

What formulas and rules did you remember in the lesson?

Analyze your work in the lesson.

The work of students in the lesson is assessed.

NS. Homework... К: Р IV (repeat) Art. 156-157 No. 4 (a-c), No. 7 (a-c),

Optional: No. 16

Application

Evaluation paper

F / I / student __________________________________________

	Crossword	Warm up	Work in notebooks	Equations	Check yourself (s \ r)

Card number 1

1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3

Decoder

Card number 2

1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3

Decoder

Card number 3

1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3

Decoder

Card number 1

1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3

Decoder

Card number 2

1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3

Decoder

Card number 3

1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3

Decoder

Card number 1

1) X 1 \ 3 = 4; 2) y -1 = 3 \ 5; 3) a 1 \ 2 = 2 \ 3; 4) x -0.5 x 1.5 = 1; 5) y 1 \ 3 = 2; 6) a 2 \ 7 and 12 \ 7 = 25; 7) a 1 \ 2: a = 1 \ 3

Decoder

Card number 2

1) X 1 \ 3 = 4; 2) y -1 = 3; 3) (x + 6) 1 \ 2 = 3; 4) y 1 \ 3 = 2; 5) (y-3) 1 \ 3 = 2; 6) a 1 \ 2: a = 1 \ 3

Decoder

Card number 3

1) a 2 \ 7 and 12 \ 7 = 25; 2) (x-12) 1 \ 3 = 2; 3) x -0.7 x 3.7 = 8; 4) a 1 \ 2: a = 1 \ 3; 5) a 1 \ 2 = 2 \ 3

Decoder

Option 1 1. Calculate: (1 point) 2. Simplify the expression: 1 point each a) x 1 \ 2 x 3 \ 4 b) (x -5 \ 6) -2 \ 3 c) x -1 \\ 3: x 3/4 d) (0.04x 7 \\ 8) -1 \\ 2 3. Solve the equation: (2 points) 4. Simplify the expression: (2 points) (a + 3a 1 \ 2): (a 1 \ 2 +3) 5. Find the value of the expression: (3 points) (Y 1 \ 2 -2) -1 - (Y 1 \ 2 +2) -1 at y = 18

Option 2 1. Calculate: (1 point) 2. Simplify the expression: 1 point each a) x 1.6 x 0.4 b) (x 3 \ 8) -5 \ 6 c) x 3 \ 7: x -2 \ 3 d) (0.008x -6 \ 7) -1 \ 3 3. Solve the equation: (2 points) 4. Simplify the expression: (2 points) (in 1.5 s - sun 1.5): (in 0.5 - s 0.5) 5. Find the value of the expression: (3 points) (x 3 \ 2 + x 1 \ 2): (x 3 \ 2 -x 1 \ 2) at x = 0.75

Sections: Mathematics

Class: 9

PURPOSE: To consolidate and improve the skills of applying the properties of the degree with a rational indicator; develop the skills of performing the simplest transformations of expressions containing powers with a fractional exponent.

LESSON TYPE: a lesson in consolidating and applying knowledge on this topic.

TEXTBOOK: Algebra 9 ed. S.A. Telyakovsky.

DURING THE CLASSES

Introductory speech of the teacher

"People unfamiliar with algebra cannot imagine the amazing things that can be achieved ... with the help of the named science." G.V. Leibniz

Algebra opens the doors to the laboratory complex for us “Degree with a rational indicator”.

1. Frontal poll

1) Give a definition of the degree with a fractional exponent.

2) For what fractional exponent is the degree defined with the base equal to zero?

3) Will there be a degree with a fractional exponent for a negative base?

Assignment: Present the number 64 as a power with a base - 2; 2; eight.

What number is 64?

Is there any other way to represent 64 as a power with a rational exponent?

2. Work in groups

1 group. Prove that expressions (-2) 3/4; 0 -2 are meaningless.

Group 2. Imagine the exponent with a fractional root: 2 2/3; 3 -1 | 3; -in 1.5; 5a 1/2; (x-y) 2/3.

Group 3. Present as a power with a fractional exponent: v3; 8 va 4; 3v2 -2; v (x + y) 2/3; vvv.

3. Let's go to the laboratory "Action on degrees"

Frequent guests of the laboratory are astronomers. They bring their "astronomical numbers", subject them to algebraic processing and get useful results.

For example, the distance from Earth to the Andromeda nebula is expressed by the number

95000000000000000000 = 95 10 18 km;

it's called quintillion.

The mass of the sun in grams is expressed by the number 1983 10 30 g - nonalion.

In addition, other serious tasks fall into the laboratory. For example, the problem of evaluating expressions like this often arises:

but) ; b); in) .

Laboratory staff perform such calculations in the most convenient way.

You can connect to work. To do this, we repeat the properties of degrees with rational exponents:

Now evaluate or simplify the expression using the properties of the rational exponents:

1st group:

Group 2:

Group 3:

Check: one person from the group at the blackboard.

4. Assignment for comparison

How do you compare the expressions 2 100 and 10 30 using the power properties?

Answer:

2 100 =(2 10) 10 =1024 10 .

10 30 =(10 3) 10 =1000 10

1024 10 >1000 10

2 100 >10 30

5. And now I invite you to the Laboratory for Research Degrees.

What transformations can we perform on degrees?

1) Present the number 3 as a power with exponent 2; 3; -one.

2) In what way can the expression a-b be factorized; in + in 1/2; a-2a 1/2; 2 x 2?

3) Reduce the fraction followed by cross-checking:

4) Explain the transformations performed and find the meaning of the expression:

6. Working with the textbook. No. 611 (d, d, f).

Group 1: (d).

Group 2: (e).

Group 3: (e).

No. 629 (a, b).

Mutual verification.

7. We carry out a workshop (independent work).

Expressions are given:

Which fractions are reduced by the formulas abbreviated multiplication and factoring out the common factor?

Group 1: No. 1, 2, 3.

Group 2: No. 4, 5, 6.

Group 3: No. 7, 8, 9.

When completing the task, you can use the recommendations.

1. If the example record contains both degrees with a rational exponent and roots of the nth degrees, then write down the roots nth degree in the form of degrees with a rational exponent.
2. Try to simplify the expression you are performing on: expanding parentheses, applying the abbreviated multiplication formula, moving from a power with a negative exponent to an expression containing exponents with a positive exponent.
3. Determine the order of actions.
4. Follow the steps in the correct order.

The teacher evaluates by collecting notebooks.

8. Homework: No. 624, 623.

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 obtain 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 hold:

• 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 containing powers with fractional exponents.

Examples of transformations of expressions containing a power 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 powers with fractional exponents.

Expressions, expression conversion

Power expressions (expressions with powers) and their conversion

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

Page navigation.

What are exponential expressions?

The term "exponential expressions" is almost never found school textbooks mathematics, but he often appears in collections of problems, especially those designed to prepare for the exam and the OGE, for example,. After analyzing the tasks in which you need to perform any actions with exponential expressions, it becomes clear that expres- sions are understood as expressions containing degrees in their records. Therefore, for yourself, you can accept the following definition:

Definition.

Power expressions Are expressions containing degrees.

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

As you know, first there is an acquaintance 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 −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 high school, they return to degrees again. There, a degree with a rational exponent is introduced, which entails the appearance of the corresponding power expressions: , , etc. Finally, degrees with irrational indicators and expressions containing them are considered:,.

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

So, we figured out the question of what are exponential expressions. Next, we will learn how to transform them.

Basic types of transformations of power expressions

With exponential expressions, you can perform any of the basic identical transformations of expressions. For example, you can expand parentheses, replace numeric expressions with their values, provide similar terms, etc. Naturally, in this case it is necessary to follow the accepted procedure for performing actions. Here are some examples.

Example.

Evaluate the value of the exponential expression 2 3 · (4 2 −12).

Solution.

According to the order of performing 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, replace the power 2 3 with its value 8, and then 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 similar terms 3 · a 4 · b −7 and 2 · a 4 · b −7, and we can bring them:.

Answer:

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

Example.

Imagine an expression with powers as a product.

Solution.

To cope with the task, the representation of the number 9 in the form of a power of 3 2 and the subsequent use of the formula for abbreviated multiplication is the difference of squares:

Answer:

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

Working with base and exponent

There are degrees, the base and / or exponent of which are not just numbers or variables, but some expressions. As an example, we present the entries (2 + 0.37) 5-3.7 and (a (a + 1) -a 2) 2 (x + 1).

When working with such expressions, you can replace both the expression based on the degree and the expression in the exponent with an identically equal expression on the ODZ of its variables. In other words, we can, according to the rules known to us, 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 above exponential expression (2 + 0.3 · 7) 5-3.7, you can perform actions with the numbers in the base and exponent, which will allow you to go to the power 4.1 1.3. And after opening the parentheses and reducing similar terms in the base of the degree (a (a + 1) −a 2) 2 (x + 1), we get a power expression more simple kind a 2 (x + 1).

Using degree properties

One of the main tools for converting expres- sions with powers is equalities, reflecting. Let us recall the main ones. For any positive numbers a and b and arbitrary real numbers r and s, the following power properties are valid:

• 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 also 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 attention when transforming power expressions is focused precisely on the ability to choose a suitable property and apply it correctly. In this case, the bases of degrees are usually positive, which allows using the properties of degrees without restrictions. The same applies to the transformation of expressions containing variables in the bases of degrees - the area permissible values of variables is usually such that the bases on it take only positive values, which allows you to freely use the properties of the degrees. In general, you need to constantly ask yourself whether it is possible in this case to apply any property of degrees, because inaccurate use of properties can lead to a narrowing of the ODV and other troubles. These points are discussed in detail and with examples in the article on conversion of expressions using degree properties. Here we restrict ourselves to a few simple examples.

Example.

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

Solution.

First, the second factor (a 2) −3 is transformed by the property of raising a power to a power: (a 2) −3 = a 2 (−3) = a −6... The original exponential expression will then 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.

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

Example.

Find the value of the exponential 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 degrees with the same bases, the indicators add up: .

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

Answer:

.

Example.

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

Solution.

The degree a 1.5 can be represented as a 0.5 · 3 and further, 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 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 be such fractions. 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 canceled, reduced to a new denominator, worked separately with their numerator and separately with the denominator, etc. To illustrate the words spoken, consider the solutions of several examples.

Example.

Simplify exponential expression .

Solution.

This exponential 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 the powers, and in the denominator we give 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 in the same way as reducing to a new denominator rational fractions... 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 ODV. 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.

Reduce 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 which additional factor helps to achieve the desired result. This is a factor of a 0.3, since a 0.7 · a 0.3 = a 0.7 + 0.3 = a. Note that on the range of permissible 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, you can find that

and multiplying this expression by will give the sum of the 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. On the range of valid 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:

but) , b) .

The abbreviation of fractions containing powers is also nothing new: the numerator and denominator are represented as a number of factors, and the same factors of the numerator and denominator are canceled.

Example.

Reduce the fraction: a) , b).

Solution.

a) First, the numerator and denominator can be reduced by the numbers 30 and 45, which is 15. Also, obviously, one can perform a reduction 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 will have to perform preliminary transformations. In this case, they consist in factoring the denominator into factors according to the formula for the difference of squares:

Answer:

but)

b) .

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

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 cancel by a power of x 1/2, after which we have .

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

Answer:

Example.

Simplify exponential expression .

Solution.

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

Answer:

.

And we also add that it is possible and in many cases desirable to transfer multipliers 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, an exponential expression can be replaced with.

Converting expressions with roots and powers

Often in expressions in which some transformations are required, along with powers with fractional exponents, there are also roots. To transform such an expression to the desired form, in most cases it is enough to go only to the roots or only to the powers. But since it is more convenient to work with degrees, they usually go from roots to degrees. However, it is advisable to carry out such a transition when the ODV 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 ODV into several intervals (we analyzed this in detail in the article the transition from roots to powers and back. a degree with an irrational indicator is introduced, which makes it possible to talk about a degree with an arbitrary real indicator. exponential function , which is analytically set by the degree, at the base of which is the number, and in the indicator - the variable. So we are faced with exponential expressions containing numbers in the base of the degree, and in the exponent - expressions with variables, and naturally it becomes necessary to perform transformations of such expressions.

It should be said that the transformation of expressions of this type usually has to be performed when solving exponential equations and exponential inequalities and these conversions are pretty simple. In the overwhelming majority of cases, they are based on the properties of the degree and are mainly aimed at introducing a new variable in the future. We can demonstrate them by the equation 5 2 x + 1 −3 5 x 7 x −14 7 2 x − 1 = 0.

First, the degrees in which the sum of a variable (or expressions 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.

Further, both sides of the equality are divided by the expression 7 2 x, which takes only positive values ​​on the ODZ of the 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 the subsequent transformations of expressions with powers ):

Fractions with powers are now canceled, which gives .

Finally, the ratio of degrees with the same exponents is replaced by the degrees of relations, which leads to the equation which is equivalent to ... The performed transformations allow introducing a new variable, which reduces the solution of the original exponential equation to the solution of the quadratic equation

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