Converting expressions containing powers. Converting Expressions

The arithmetic operation that is performed last when calculating the value of an expression is the “master” operation.

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 factorized).

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

To reinforce this, solve a few examples yourself:

Examples:

Solutions:

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

The first step should be factorization:

4. Adding and subtracting fractions. Reducing fractions to a common denominator.

Adding and subtracting ordinary fractions is a 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 relatively prime, 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 convert mixed fractions into improper ones, and then according to the usual scheme:

It's a completely different matter if the fractions contain letters, for example:

Let's start with something simple:

a) Denominators do not contain letters

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

Now in the numerator you can give 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 one at a time;

· and multiply them by all other non-common factors.

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

Let us emphasize the common factors:

Now let’s write out the common factors one at a time and add to them all the 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:

· factor the denominators;

· determine common (identical) factors;

· write out all common factors once;

· multiply them by all other non-common factors.

So, in order:

1) factor the denominators:

2) determine common (identical) factors:

3) write out all common factors once and multiply them by all other (unemphasized) factors:

So there's a common denominator 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 a degree

to a degree

to a degree

to a degree.

Let's complicate the task:

How to make fractions have the same denominator?

Let's remember the basic property of a fraction:

Nowhere does it say 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 did you learn?

So, another unshakable rule:

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

But what do you need to multiply by to get?

So multiply by. And multiply by:

We will call expressions that cannot be factorized “elementary factors.”

For example, - this is an elementary factor. - Same. But no: it can be factorized.

What about the 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 deal with them in the same way.

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

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

Another example:

Solution:

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

Great! Then:

Another example:

Solution:

As usual, let's 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 similar... And it’s true:

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 let's bring it to a common denominator:

Got it? Let's check it now.

Tasks for independent solution:

Answers:

Here we need to 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 double product. The partial square of the sum is one of the factors in the expansion of the difference of cubes:

What to do if there are already three fractions?

Yes, the same thing! First of all, let’s make sure that the maximum number of factors in the denominators is the same:

Please note: 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 changes again to the opposite. As a result, it (the sign in front of the fraction) has not changed.

We write out the entire first denominator 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... It’s clear what to do with fractions. But what about the two?

It's simple: you know how to add fractions, right? So, we need to make two become a fraction! Let's remember: a fraction is a division operation (the numerator is divided by the denominator, in case you 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 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 numerical expression? Remember by calculating the meaning of this expression:

Did you count?

It should work.

So, let me remind you.

The first step is to calculate the degree.

The second is multiplication and division. If there are several multiplications and divisions at the same time, they can be done in any order.

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

But: the expression in brackets is evaluated out of turn!

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

What if there are more brackets inside the brackets? Well, let's think: some expression is written inside the brackets. When calculating an expression, what should you do first? That's right, calculate the brackets. 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, adding fractions, reducing fractions, and so on. The only difference will be the action of factoring polynomials (we often use this when working with fractions). Most often, to factorize, you need to use I or simply put the common factor out of brackets.

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

For example:

Let's simplify the expression.

1) First, we simplify the expression in brackets. There we have a 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 any further; all the factors here are elementary (do you still remember what this means?).

2) We get:

Multiplying fractions: what could be simpler.

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 determine the order of actions.

First, let's add the fractions in parentheses, so instead of two fractions we get one.

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

I will number the steps schematically:

Now I’ll show you the process, tinting the current action in red:

1. If there are similar ones, they must be brought immediately. At whatever point similar ones arise in our country, it is advisable to bring them up immediately.

2. The same applies to reducing fractions: as soon as the opportunity to reduce appears, it must be taken advantage of. The exception is for 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 what was promised at the very beginning:

Answers:

Solutions (brief):

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

Now on to learning!

CONVERTING EXPRESSIONS. SUMMARY AND BASIC FORMULAS

Basic simplification operations:

• Bringing similar: to add (reduce) similar terms, you need to add their coefficients and assign the letter part.
• Factorization: putting the common factor out of brackets, applying it, etc.
• Reducing a fraction: 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 factorize
  2) if the numerator and denominator have common factors, they can be crossed out.

  IMPORTANT: only multipliers can be reduced!

• Adding and subtracting fractions:
  ;
• Multiplying and dividing fractions:
  ;

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.

Page navigation.

What are power expressions?

The term “power expressions” practically does not appear 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 degrees.

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

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 negative integer 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 transform 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 of identical 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 of a simpler form 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 the same bases, the exponents 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 which additional multiplier helps to achieve the 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 transform 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 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 of the original exponential equation to the solution of a quadratic equation

Municipal state educational institution

basic secondary school No. 25

Algebra lesson

Subject:

« Converting expressions containing powers with fractional exponents"

Developed by:

,

mathematic teacher

higher toqualification category

Nodal

2013

Lesson topic: Converting expressions containing exponents with fractional exponents

The purpose of the lesson:

1. Further development of skills, knowledge, and skills in converting expressions containing degrees with fractional exponents

2. Development of the ability to find errors, development of thinking, creativity, speech, computing skills

3. Fostering independence, interest in the subject, attentiveness, accuracy.

TCO: magnetic board, test cards, tables, individual cards, schoolchildren have blank signed sheets on the table for individual work, a crossword puzzle, tables for mathematical warm-up, a multimedia projector.

Lesson type: securing ZUN.

Lesson plan over time

1. Organizational aspects (2 min)

2. Checking homework (5 min)

3. Crossword puzzle (3 min)

4. Mathematical warm-up (5 min)

5. Solving frontal strengthening exercises (7 min)

6. Individual work (10 min)

7. Solution of repetition exercises (5 min)

8. Lesson summary (2 min)

9. Homework assignment (1 min)

During the classes

1) Checking homework in the form of peer review . Good students check the notebooks of weak children. And the weak guys check with the strong ones using a sample control card. Homework is given in two versions.

I option the task is not difficult

II option the task is difficult

As a result of the check, the guys highlight the mistakes with a simple pencil and give a rating. I finally check the work after the children hand in their notebooks after class. I ask the guys the results of their test and put grades for this type of work in my summary table.

2) To test theoretical material, a crossword puzzle is offered.

Vertically:

1. Property of multiplication used when multiplying a monomial by a polynomial?

2. The effect of exponents when raising a power to a power?

3. A degree with zero index?

4. A product consisting of identical factors?

Horizontally:

5. Root n – oh degree of a non-negative number?

6. The action of exponents when multiplying powers?

7. The effect of exponents in dividing powers?

8. The number of all identical factors?

3) Mathematical warm-up

a) perform the calculation and use the cipher to read the word hidden in the problem.

There is a table on the board in front of you. The table in column 1 contains examples that need to be calculated.

Key to table

And write the answer in the column II, and in column III put the letter corresponding to this answer.

Teacher: So, the encrypted word is “degree”. In the next task we work with the 2nd and 3rd degrees

b) Game “Make sure you don’t make a mistake”

Instead of dots, put a number

a) x=(x...)2; b) a3/2 = (a1/2)…; c) a=(a1/3)…; d) 5... = (51/4)2; e) 34/3=(34/9)…; e) 74/5 = (7...)2; g) x1/2=(x...)2; h) y1/2=(y...)2

Let's find the error:

А1/4 – 2а1/2 + 1 = (а1/

So, guys, what needed to be used to complete this task:

Property of degrees: when raising a degree to a power, the exponents are multiplied;

4) Now let’s get started with the front-end written work. , using the results of previous work. Open notebooks and write down the date and topic of the lesson.

№ 000

a) a – b = (a1/2)2 – (b1/2)2 = (a1/2 – b1/2)*(a1/2 + b1/2)

b) a – c = (a1/3)3 – (b1/3)3 = (a1/3 – c1/3)*(a2/3 + a1/3 b1/3 + b2/3)

No. 000 (a, c, d, e)

A ) m2 – 5 = m2 – (m1/2)2 = (m – 51/2)*(m+51/2)

c) a3 – 4 = (a3/2)2 – 22 = (a3/2 – 2)*(a3/2 +2)

d) x2/5 – y4/5 = (x1/5)2 – (y2/5)2 = (x1/5 – y2/5)*(x1/5 + y2/5)

e) 4 – a = 22 – (a1/2)2 = (2 – a1/2)*(2+a1/2)

No. 000 (a, d, f)

a) x3 – 2 = x3 – (21/3)3 = (x – 21/3)*(x2 + 21/3 x + 22/3)

d) a6/5 + 27 = (a2/5)3 + 33 = (a2/5 + 3)*(a4/3 – 3 a2/5 + 9)

e) 4 + y = (41/3)3 + (y1/3)3 = (41/3 + y1/3)*(42/3 + 41/3 y1/3 + y2/3)

Grade

5) Work on individual cards using four options on separate sheets

Tasks with varying degrees of difficulty are completed without any prompting from the teacher.

I check the work immediately and put grades in my table and on the guys’ sheets.

No. 000 (a, c, d, h)

a) 4*31/2/(31/2 – 3) = 4*31/2 /31/2*(1 – 31/2) = 4 / (1 – 31/2)

c) x + x1/2 /2x = x1/2*(x1/2+1)/ 2*(x1/2)2 = (x1/2+1)/ 2x1/2

e) (a2/3 – b2/3)/(a1/3 +b1/3) = (a1/3)2 – (b1/3)2/(a1/3 +b1/3) = (a1/3 + b1/3)*(a1/3 – b1/3)/(a1/3 + b1/3) = a1/3 – b1/3

h) (x2/3 - x1/3 y1/3 + y2/3)/(x + y) = ((x1/3)2 – x1/3 y1/3 + (y1/3)2)/(( x1/3)3 +(y1/3)3) = ((x1/3)2 – x1/3 y1/3 +(y1/3)2)/(x1/3 +y1/3)*((x1 /3)2 – x1/3 y1/3 + (y1/3)2) = 1/ (x1/3 + y1/3)

7) Working on individual cards with varying degrees of complexity. In some exercises there are recommendations from the teacher, since the material is complicated and it is difficult for weak children to cope with the work

There are also four options available. The assessment takes place immediately. I put all the grades in a spreadsheet.

Problem No. from the collection

The teacher asks questions:

1. What should be found in the problem?

2. What do you need to know for this?

3. How to express the time of 1 pedestrian and 2 pedestrians?

4. Compare the times of pedestrians 1 and 2 according to the conditions of the problem and create an equation.

The solution of the problem:

Let x (km/h) be the speed of 1 pedestrian

X +1 (km/h) – speed of 2 pedestrians

4/х (h) – pedestrian time

4/(x +1) (h) – time of the second pedestrian

According to the conditions of the problem 4/x >4/ (x +1) for 12 minutes

12 min = 12 /60 h = 1/5 h

Let's make an equation

X/4 – 4/ (x +1) = 1/5

NOZ: 5x(x +1) ≠ 0

5*4*(x+1) – 5*4x = x*(x+1)

20x + 20 – 20x – x2 – x = 0

X2 +x –20 = 0

D=1 – 4*(-20) = 81, 81>0, 2 k

x1 = (-1 -√81)/(-2) = 5 km/h – speed of 1 pedestrian

x2 = (-1 + √81)/(-2) = 4 – does not fit the meaning of the problem, since x>0

Answer: 5 km/h – speed of 2 pedestrians

9) Lesson summary: So, guys, today in the lesson we consolidated knowledge, skills, and skills of transforming expressions containing degrees, applied abbreviated multiplication formulas, moved the common factor out of brackets, and repeated the material covered. I point out the advantages and disadvantages.

Summarizing the lesson in a table.

10) I announce the grades. Homework assignment

Individual cards K – 1 and K – 2

I change B – 1 and B – 2; B – 3 and B – 4, since they are equivalent

Applications to the lesson.

1) Cards for homework

1. simplify

a) (x1/2 – y1/2)2 + 2x1/2 y1/2

b) (a3/2 + 5a1\2)2 – 10a2

2. present as a sum

a) a1/3 c1\4*(b2/3 + c3/4)

b) (a1/2 – b1/2)*(a + a1/2 b1\2 + c)

3. take out the overall multiplier

c) 151/3 +201/3

1. simplify

a) √m + √n – (m1/4 – n1/4)2

b) (a1/4 + b1/4)*(a1/8 + b1/8)*(a1\8 – b1/8)

2. present as a sum

a) x0.5 y0.5*(x-0.5 – y1.5)

b) (x1/3 + y1/3)*(x2\3 – x1/3 y1\3 + y2/3)

3. Take the common factor out of brackets

b) c1\3 – c

c) (2a)1/3 – (5a)1\3

2) control card for B – 2

a) √m + √n – (m 1|4 – n 1|4)2 = m 1|2 + n 1|2 – ((m 1|2)2 – 2 m 1/4 n 1/4 + (n 1/2)2) = m 1/2 + n 1/2 – m 1/2 + 2 m 1/4 n 1/4 – n 1/2 = 2 m 1/4 n 1/4

b) (a1/4 + b1/4)*(a1/8 + b1/8)*(a1/8 – b1/8) = (a1/4 + b1/4)*(a1/8)2 – ( в1/8)2 = (а1/4 + в1/4)*(а1/4 – в1/4) = (а1/4)2 – (в1/4)2 = а1/2 – в1/2

a) x0.5 y0.5* (x-0.5-y1.5) = x0.5 y0.5 x-0.5 – x0.5 y0.5y1.5 = x0 y0.5 – x0.5 y2 = y0.5 – x0.5 y2

b) (x1/3 + y1/3)*(x2/3 – x1/3 y1\3 + y2/3) = (x1\3 + y1/3)*((x1/3)2 – x1/3 y1\3 + (y1/3)2) = (x1/3)2 + (y1/3)2 = x + y

a) 3 – 31/2 = 31/2 * (31/2 - 1)

b) v1/3 – v = v1/3 *(1 – v2/3)

c) (2a)1/3 – (5a)1/3 = a1/3*(21/3 – 51/3)

3) Cards for the first individual work

a) a – y, x ≥ 0, y ≥ 0

b) a – and, a ≥ 0

1. Factorize as a difference of squares

a) a1/2 – b1/2

2. Factorize as a difference or sum of cubes

a) c1/3 + d1/3

1. Factorize as a difference of squares

a) X1/2 + Y1/2

b) X1/4 – U1/4

2. Factorize as a difference or sum of cubes

4) cards for the second individual work

a) (x – x1/2)/ (x1/2 – 1)

Instruction: x1/2, remove numerators from brackets

b) (a - c)/(a1/2 – b1/2)

Note: a – b = (a1/2)2 – (b1/2)2

Reduce the fraction

a) (21/4 – 2)/ 5*21/4

Instruction: remove 21/4 from brackets

b) (a – c)/(5а1/2 – 5в1/2)

Note: a – b = (a1/2)2 – (b1/2)2

Option 3

1. Reduce the fraction

a) (x1/2 – x1/4)/x3/4

Instruction: place x1/4 out of brackets

b) (a1/2 – b1/2)/(4a1/4 – 4b1/4)

Option 4

Reduce the fraction

a) 10/ (10 – 101/2)

b) (a - c)/(a2/3 + a1\3b1/3+ B 1/3)

Subject: " Converting expressions containing powers with a fractional exponent"

“Let someone try to eliminate degrees from mathematics, and he will see that without them you won’t get far.” (M.V. Lomonosov)

Lesson objectives:

educational: summarize and systematize students’ knowledge on the topic “Degree with a rational indicator”; monitor the level of mastery of the material; eliminate gaps in students’ knowledge and skills;

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

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

Lesson type: lesson of generalization and systematization of knowledge

Equipment: assessment sheets, cards with tasks, decoders, crossword puzzles for each student.

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

DURING THE CLASSES

I. Organizational moment.

Teacher: We have finished studying the topic “A power with a rational exponent and its properties.” Your task in this lesson is to show how you have mastered the material you have studied and how you can apply the acquired knowledge to solve specific problems. Each of you has a score sheet on your desk. In it you will enter your assessment for each stage of the lesson. At the end of the lesson you will give an average score for the lesson.

Evaluation paper

II. Checking homework.

Peer checking with a pencil in hand, the answers are read out by the students.

III. Updating students' knowledge.

Teacher: The famous French writer Anatole France once said: “Learning must be fun... To absorb knowledge, you must absorb it with appetite.”

Let's repeat the necessary theoretical information while solving the crossword puzzle.

Horizontally:

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

2. Product consisting of identical factors (degree).

3. The action of exponents when raising a power to a power (work).

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

Vertically:

5. Number of all identical factors (index).

6. Degree with zero index (unit).

7. Repeating multiplier (base).

8. Value of 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 a degree with a rational exponent and its properties, and complete the following tasks.

1. Present the expression x 22 as a product of two powers with a base x, if one of the factors is equal to: 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) from 1.4 from -0.3 from 2.9

3. Calculate and compose the word using a decoder.

After completing this task, you guys will find out 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 (Stifel)

V. Written work in notebooks (answers are opened 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. Work in groups.

Exercise. Solve equations and form words using a decoder.

Card No. 1

Word: 1234567 (Diophantus)

Card No. 2

Card No. 3

Word: 123451 (Newton)

Decoder

Teacher. All these scientists contributed to the development of the concept of “degree”.

VII. Historical information about the development of the concept of degree (student message).

The concept of a degree with a natural indicator was formed among ancient peoples. Square and cube numbers were used to calculate areas and volumes. The powers of some numbers were used in solving certain problems by scientists of Ancient Egypt and Babylon.

In the 3rd century, the book of the Greek scientist Diophantus “Arithmetic” was published, which laid the foundation for the introduction of letter symbols. Diophantus introduces symbols for the first six powers of the unknown and their reciprocals. In this book, a square is denoted by a sign with a subscript r; cube – sign k with index r, etc.

From the practice of solving more complex algebraic problems and operating with degrees, the need arose to generalize the concept of degree and expand it by introducing zero, negative and fractional numbers as an exponent. Mathematicians came to the idea of ​​generalizing the concept of degree to a degree with a non-natural exponent gradually.

Fractional exponents and the simplest rules for operating powers with fractional exponents are found in the French mathematician Nicholas Oresme (1323–1382) in his work “Algorithm of Proportions.”

The equality, a 0 =1 (for and not equal to 0) was used in his works at the beginning of the 15th century by the Samarkand scientist Giyasaddin Kashi Dzhemshid. Independently, the zero indicator was introduced by Nikolai Schuke in the 15th century. It is known that Nicholas Shuquet (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 that a 1/n is meant to be a root.

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

At the end of the 16th century, François Viète introduced letters to designate not only variables, but also their coefficients. He used abbreviations: N, Q, C - for the first, second and third degrees. But modern notations (such as a 4, a 5) were introduced in the 17th century by Rene Descartes.

Modern definitions and notations for powers with zero, negative and fractional exponents originate from the work of English mathematicians John Wallis (1616–1703) and Isaac Newton (1643–1727).

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

The introduction of a degree with a rational exponent is one of many examples of generalization of the concepts of mathematical action. A degree with zero, negative and fractional exponents is defined in such a way that the same rules of action are applied to it as for a degree with a natural exponent, i.e. so that the basic properties of the original defined 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 the new definition of a degree with a rational exponent remains the same for the special 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, and it was fully and clearly established by the German mathematician G. Hankel in 1867.

VIII. Check yourself.

Independent work using cards (answers are revealed 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 class?

Analyze your work in class.

Students' work in class is assessed.

X. Homework. K: R IV (repeat) art. 156-157 No. 4 (a-c), No. 7 (a-c),

Additional: No. 16

Application

Evaluation paper

Name/name/student_______________________________________________

Card No. 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 a 12\7 = 25; 7) a 1\2: a = 1\3

Decoder

Card No. 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 No. 3

1) a 2\7 a 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) and 1\2 = 2\3

Decoder

Card No. 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 a 12\7 = 25; 7) a 1\2: a = 1\3

Decoder

Card No. 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 No. 3

1) a 2\7 a 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) and 1\2 = 2\3

Decoder

Card No. 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 a 12\7 = 25; 7) a 1\2: a = 1\3

Decoder

Card No. 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 No. 3

1) a 2\7 a 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) and 1\2 = 2\3

Decoder

Sections: Mathematics

Class: 9

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

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

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

DURING THE CLASSES

Teacher's opening speech

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

Algebra opens the doors to the laboratory complex for us “A degree with a rational exponent.”

1. Frontal survey

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

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

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

Assignment: Imagine the number 64 as a power with base - 2; 2; 8.

The cube of what number is 64?

Is there another way to represent the number 64 as a power with a rational exponent?

2. Work in groups

1 group. Prove that the expressions (-2) 3/4 ; 0 -2 doesn't make sense.

2nd group. Imagine a power with a fractional exponent in the form of a root: 2 2/3; 3 -1|3 ; -in 1.5; 5a 1/2; (x-y) 2/3 .

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

3. Let’s move on to the laboratory “Action on powers”

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 - nonnalion.

In addition, the laboratory faces other serious tasks. For example, the problem of calculating expressions like:

A) ; b) ; V) .

Laboratory staff perform such calculations in the most convenient way.

You can connect to work. To do this, let us repeat the properties of powers with rational exponents:

Now calculate or simplify the expression using the properties of powers with rational exponents:

1st group:

Group 2:

Group 3:

Check: one person from the group at the board.

4. Comparison task

How can we compare the expressions 2 100 and 10 30 using the properties of powers?

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 “Research of Degrees” laboratory.

What transformations can we perform on powers?

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

2) How can expressions a-c be factorized? in+in 1/2; a-2a 1/2; 2's 2?

3) Reduce the fraction followed by mutual verification:

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

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

Group 1: (d).

Group 2: (e).

Group 3: (f).

No. 629 (a, b).

Peer review.

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

Given expressions:

When reducing which fractions are abbreviated multiplication formulas and putting the common factor out of brackets?

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 recommendations.

1. If the example notation contains both powers with a rational exponent and roots of the nth degree, then write the roots of the nth degree in the form of powers with a rational exponent.
2. Try to simplify the expression on which the actions are performed: opening parentheses, using the abbreviated multiplication formula, moving from a power with a negative exponent to an expression containing powers with a positive exponent.
3. Determine the order in which the actions should be performed.
4. Complete the steps in the order in which they are performed.

The teacher evaluates after collecting the notebooks.

8. Homework: No. 624, 623.