Basic information about rational expressions and their transformations. Transformation of rational expressions Fractional rational expressions examples with solutions

First of all, in order to learn how to work with rational fractions without errors, you need to learn the abbreviated multiplication formulas. And it's not easy to learn - they need to be recognized even when sines, logarithms and roots act as terms.

However, the main tool remains the factorization of the numerator and denominator of a rational fraction. This can be accomplished in three different ways:

Actually, according to the formula of abbreviated multiplication: they allow you to fold a polynomial into one or more factors;
Using the factorization of a square trinomial in terms of the discriminant. The same method allows us to make sure that any trinomial does not decompose into factors at all;
The grouping method is the most difficult tool, but it is the only method that works if the previous two did not work.

As you've probably guessed from the title of this video, we're going to talk about rational fractions again. Just a few minutes ago, I finished a lesson with one tenth grader, and there we were analyzing these very expressions. Therefore, this lesson will be intended specifically for high school students.

Surely many will now have a question: "Why should pupils in grades 10-11 learn such simple things as rational fractions, because this is done in grade 8?" But the trouble is that most people just "pass" this topic. In grades 10-11, they no longer remember how multiplication, division, subtraction and addition of rational fractions from the 8th grade is done, but it is on this simple knowledge that further, more complex constructions are built, such as the solution of logarithmic, trigonometric equations and many others complex expressions, so there is practically nothing to do in high school without rational fractions.

Formulas for solving problems

Let's get down to business. First of all, we need two facts - two sets of formulas. First of all, you need to know the abbreviated multiplication formulas:

$ ((a) ^ (2)) - ((b) ^ (2)) = \ left (a-b \ right) \ left (a + b \ right) $ - difference of squares;
$ ((a) ^ (2)) \ pm 2ab + ((b) ^ (2)) = ((\ left (a \ pm b \ right)) ^ (2)) $ - the square of the sum or difference;
$ ((a) ^ (3)) + ((b) ^ (3)) = \ left (a + b \ right) \ left (((a) ^ (2)) - ab + ((b) ^ ( 2)) \ right) $ - the sum of cubes;
$ ((a) ^ (3)) - ((b) ^ (3)) = \ left (ab \ right) \ left (((a) ^ (2)) + ab + ((b) ^ (2) ) \ right) $ - difference of cubes.

In their pure form, they are not found in any examples and in real serious expressions. Therefore, our task is to learn to see much more complex constructions under the letters $ a $ and $ b $, for example, logarithms, roots, sines, etc. You can learn to see this only through constant practice. That is why solving rational fractions is absolutely necessary.

The second, completely obvious formula is the factorization of a square trinomial:

$ ((x) _ (1)) $; $ ((x) _ (2)) $ - roots.

We have dealt with the theoretical part. But how to solve real rational fractions, which are considered in grade 8? Now we are going to practice.

Problem number 1

\ [\ frac (27 ((a) ^ (3)) - 64 ((b) ^ (3))) (((b) ^ (3)) - 4): \ frac (9 ((a) ^ (2)) + 12ab + 16 ((b) ^ (2))) (((b) ^ (2)) + 4b + 4) \]

Let's try to apply the above formulas to solving rational fractions. First of all, I want to explain why factoring is needed at all. The fact is that at the first glance at the first part of the task, you want to reduce a cube with a square, but this is absolutely impossible, because they are terms in the numerator and in the denominator, but in no case are factors.

In general, what is a reduction? Abbreviation is a general rule of thumb for dealing with such expressions. The main property of a fraction is that we can multiply the numerator and denominator by the same number, other than zero. In this case, when we reduce, then, on the contrary, we divide by the same number, different from "zero". However, we must divide all the terms in the denominator by the same number. You can't do that. And we have the right to cancel the numerator with the denominator only when both of them are factorized. Let's do it.

Now you need to see how many terms are in one or another element, in accordance with this, find out which formula should be used.

Let's convert each expression to an exact cube:

Let's rewrite the numerator:

\ [((\ left (3a \ right)) ^ (3)) - ((\ left (4b \ right)) ^ (3)) = \ left (3a-4b \ right) \ left (((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2)) \ right) \]

Let's take a look at the denominator. Let's expand it according to the formula of the difference of squares:

\ [((b) ^ (2)) - 4 = ((b) ^ (2)) - ((2) ^ (2)) = \ left (b-2 \ right) \ left (b + 2 \ right) \]

Now let's look at the second part of the expression:

Numerator:

It remains to figure out the denominator:

\ [((b) ^ (2)) + 2 \ cdot 2b + ((2) ^ (2)) = ((\ left (b + 2 \ right)) ^ (2)) \]

Let's rewrite the whole construction taking into account the above facts:

\ [\ frac (\ left (3a-4b \ right) \ left (((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2 )) \ right)) (\ left (b-2 \ right) \ left (b + 2 \ right)) \ cdot \ frac (((\ left (b + 2 \ right)) ^ (2))) ( ((\ left (3a \ right)) ^ (2)) + 3a \ cdot 4b + ((\ left (4b \ right)) ^ (2))) = \]

\ [= \ frac (\ left (3a-4b \ right) \ left (b + 2 \ right)) (\ left (b-2 \ right)) \]

Nuances of multiplying rational fractions

The key takeaway from these constructions is as follows:

Not every polynomial is factorized.
Even if it unfolds, it is necessary to carefully look at which exact formula for abbreviated multiplication.

To do this, firstly, you need to estimate how many summands there are (if there are two of them, then all we can do is expand them either by the sum of the difference of squares, or by the sum or difference of cubes; and if there are three of them, then this is , unambiguously, either the square of the sum, or the square of the difference). It often happens that either the numerator or the denominator does not require factorization at all, it can be linear, or its discriminant will be negative.

Problem number 2

\ [\ frac (3-6x) (2 ((x) ^ (2)) + 4x + 8) \ cdot \ frac (2x + 1) (((x) ^ (2)) + 4-4x) \ cdot \ frac (8 - ((x) ^ (3))) (4 ((x) ^ (2)) - 1) \]

In general, the scheme for solving this problem is no different from the previous one - there will simply be more actions, and they will become more diverse.

Let's start with the first fraction: let's look at its numerator and make possible transformations:

Now let's look at the denominator:

With the second fraction: nothing at all can be done in the numerator, because this is a linear expression, and you cannot take out any factor from it. Let's look at the denominator:

\ [((x) ^ (2)) - 4x + 4 = ((x) ^ (2)) - 2 \ cdot 2x + ((2) ^ (2)) = ((\ left (x-2 \ right )) ^ (2)) \]

We go to the third fraction. Numerator:

Let's deal with the denominator of the last fraction:

Let's rewrite the expression taking into account the above facts:

\ [\ frac (3 \ left (1-2x \ right)) (2 \ left (((x) ^ (2)) + 2x + 4 \ right)) \ cdot \ frac (2x + 1) ((( \ left (x-2 \ right)) ^ (2))) \ cdot \ frac (\ left (2-x \ right) \ left (((2) ^ (2)) + 2x + ((x) ^ ( 2)) \ right)) (\ left (2x-1 \ right) \ left (2x + 1 \ right)) = \]

\ [= \ frac (-3) (2 \ left (2-x \ right)) = - \ frac (3) (2 \ left (2-x \ right)) = \ frac (3) (2 \ left (x-2 \ right)) \]

Solution nuances

As you can see, not everything and not always rests on the abbreviated multiplication formulas - sometimes it is just enough to exclude a constant or a variable from the brackets. However, there is also the opposite situation, when there are so many terms or they are constructed in such a way that the formulas for abbreviated multiplication to them are generally impossible. In this case, a universal tool comes to our aid, namely, the grouping method. This is what we will now apply in the next task.

Problem number 3

\ [\ frac (((a) ^ (2)) + ab) (5a - ((a) ^ (2)) + ((b) ^ (2)) - 5b) \ cdot \ frac (((a ) ^ (2)) - ((b) ^ (2)) + 25-10a) (((a) ^ (2)) - ((b) ^ (2))) \]

Let's take a look at the first part:

\ [((a) ^ (2)) + ab = a \ left (a + b \ right) \]

\ [= 5 \ left (ab \ right) - \ left (ab \ right) \ left (a + b \ right) = \ left (ab \ right) \ left (5-1 \ left (a + b \ right ) \ right) = \]

\ [= \ left (a-b \ right) \ left (5-a-b \ right) \]

Let's rewrite the original expression:

\ [\ frac (a \ left (a + b \ right)) (\ left (ab \ right) \ left (5-ab \ right)) \ cdot \ frac (((a) ^ (2)) - ( (b) ^ (2)) + 25-10a) (((a) ^ (2)) - ((b) ^ (2))) \]

Now let's deal with the second bracket:

\ [((a) ^ (2)) - ((b) ^ (2)) + 25-10a = ((a) ^ (2)) - 10a + 25 - ((b) ^ (2)) = \ left (((a) ^ (2)) - 2 \ cdot 5a + ((5) ^ (2)) \ right) - ((b) ^ (2)) = \]

\ [= ((\ left (a-5 \ right)) ^ (2)) - ((b) ^ (2)) = \ left (a-5-b \ right) \ left (a-5 + b \ right) \]

Since the two elements could not be grouped, we grouped three. It remains only to figure out the denominator of the last fraction:

\ [((a) ^ (2)) - ((b) ^ (2)) = \ left (a-b \ right) \ left (a + b \ right) \]

Now let's rewrite our entire construction:

\ [\ frac (a \ left (a + b \ right)) (\ left (ab \ right) \ left (5-ab \ right)) \ cdot \ frac (\ left (a-5-b \ right) \ left (a-5 + b \ right)) (\ left (ab \ right) \ left (a + b \ right)) = \ frac (a \ left (b-a + 5 \ right)) ((( \ left (ab \ right)) ^ (2))) \]

The problem is solved, and nothing more can be simplified here.

Solution nuances

We figured out the grouping and got another very powerful tool that expands the possibilities for factoring. But the problem is that in real life no one will give us such refined examples, where there are several fractions, in which you only need to factor the numerator and denominator into a factor, and then reduce them if possible. Real expressions will be much more complex.

Most likely, in addition to multiplication and division, there will be subtractions and additions, all kinds of brackets - in general, the order of actions will have to be taken into account. But the worst thing is that when subtracting and adding fractions with different denominators, they will have to be reduced to one common. To do this, each of them will need to be factorized, and then these fractions will need to be transformed: to bring similar ones and much more. How to do it correctly, quickly, and at the same time get an unambiguously correct answer? This is what we will talk about now with the example of the following construction.

Problem number 4

\ [\ left (((x) ^ (2)) + \ frac (27) (x) \ right) \ cdot \ left (\ frac (1) (x + 3) + \ frac (1) ((( x) ^ (2)) - 3x + 9) \ right) \]

Let's write out the first fraction and try to deal with it separately:

\ [((x) ^ (2)) + \ frac (27) (x) = \ frac (((x) ^ (2))) (1) + \ frac (27) (x) = \ frac ( ((x) ^ (3))) (x) + \ frac (27) (x) = \ frac (((x) ^ (3)) + 27) (x) = \ frac (((x) ^ (3)) + ((3) ^ (3))) (x) = \]

\ [= \ frac (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) (x) \]

Let's move on to the second. Let's immediately calculate the discriminant of the denominator:

It cannot be factorized, so we write the following:

\ [\ frac (1) (x + 3) + \ frac (1) (((x) ^ (2)) - 3x + 9) = \ frac (((x) ^ (2)) - 3x + 9 + x + 3) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) = \]

\ [= \ frac (((x) ^ (2)) - 2x + 12) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) \]

Let's write out the numerator separately:

\ [((x) ^ (2)) - 2x + 12 = 0 \]

Consequently, this polynomial cannot be factorized.

The maximum that we could do and expand, we have already done.

So, we rewrite our original construction and get:

\ [\ frac (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) (x) \ cdot \ frac (((x) ^ (2) ) -2x + 12) (\ left (x + 3 \ right) \ left (((x) ^ (2)) - 3x + 9 \ right)) = \ frac (((x) ^ (2)) - 2x + 12) (x) \]

That's it, the problem is solved.

To be honest, it was not such a difficult task: everything was easily decomposed into factors, such terms were quickly given, and everything was reduced beautifully. So now let's try to solve a more serious problem.

Problem number 5

\ [\ left (\ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (((x) ^ (3) ) -8) - \ frac (1) (x-2) \ right) \ cdot \ left (\ frac (((x) ^ (2))) (((x) ^ (2)) - 4) - \ frac (2) (2-x) \ right) \]

First, let's deal with the first parenthesis. From the very beginning, factor the denominator of the second fraction separately:

\ [((x) ^ (3)) - 8 = ((x) ^ (3)) - ((2) ^ (3)) = \ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right) \]

\ [\ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (((x) ^ (3)) - 8 ) - \ frac (1) (((x) ^ (2))) = \]

\ [= \ frac (x) (((x) ^ (2)) + 2x + 4) + \ frac (((x) ^ (2)) + 8) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) - \ frac (1) (x-2) = \]

\ [= \ frac (x \ left (x-2 \ right) + ((x) ^ (2)) + 8- \ left (((x) ^ (2)) + 2x + 4 \ right)) ( \ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \]

\ [= \ frac (((x) ^ (2)) - 2x + ((x) ^ (2)) + 8 - ((x) ^ (2)) - 2x-4) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \]

\ [= \ frac (((x) ^ (2)) - 4x + 4) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right)) = \ frac (((\ left (x-2 \ right)) ^ (2))) (\ left (x-2 \ right) \ left (((x) ^ (2)) + 2x + 4 \ right )) = \ frac (x-2) (((x) ^ (2)) + 2x + 4) \]

Now let's work with the second fraction:

\ [\ frac (((x) ^ (2))) (((x) ^ (2)) - 4) - \ frac (2) (2-x) = \ frac (((x) ^ (2 ))) (\ left (x-2 \ right) \ left (x + 2 \ right)) - \ frac (2) (2-x) = \ frac (((x) ^ (2)) + 2 \ left (x-2 \ right)) (\ left (x-2 \ right) \ left (x + 2 \ right)) = \]

\ [= \ frac (((x) ^ (2)) + 2x + 4) (\ left (x-2 \ right) \ left (x + 2 \ right)) \]

Go back to our original construction and write:

\ [\ frac (x-2) (((x) ^ (2)) + 2x + 4) \ cdot \ frac (((x) ^ (2)) + 2x + 4) (\ left (x-2 \ right) \ left (x + 2 \ right)) = \ frac (1) (x + 2) \]

Key points

Once again, the key facts of today's video tutorial:

It is necessary to know "by heart" the formulas of abbreviated multiplication - and not just to know, but to be able to see in those expressions that you will encounter in real problems. A wonderful rule can help us in this: if there are two terms, then this is either the difference of squares, or the difference or sum of cubes; if three, it can only be the square of the sum or difference.
If some construction cannot be decomposed using abbreviated multiplication formulas, then either the standard formula for factoring trinomials into factors, or the grouping method comes to our aid.
If something does not work out, take a close look at the original expression - and whether any conversions are required with it at all. Perhaps it will be enough just to put the factor outside the parenthesis, and this is very often just a constant.
In complex expressions where you need to perform several actions in a row, do not forget to bring to a common denominator, and only after that, when all the fractions are reduced to it, be sure to bring something like this in a new numerator, and then factor the new numerator again - it is possible that -that will decrease.

That's all I wanted to tell you today about rational fractions. If something is not clear, there are still a bunch of video tutorials on the site, as well as a bunch of tasks for independent solution. Therefore, stay with us!

In the previous lesson, the concept of rational expression was already introduced, in today's lesson we continue to work with rational expressions and focus on transforming them. Using specific examples, we will consider methods for solving problems on transforming rational expressions and proving the related identities.

Theme:Algebraic fractions. Arithmetic operations on algebraic fractions

Lesson:Converting rational expressions

Let us first recall the definition of a rational expression.

Definition.Rationalexpression- an algebraic expression that does not contain roots and includes only the actions of addition, subtraction, multiplication and division (raising to a power).

By the concept of "transform a rational expression" we mean, first of all, its simplification. And this is carried out in the order of actions known to us: first the actions in brackets, then product of numbers(exponentiation), division of numbers, and then addition / subtraction actions.

The main goal of today's lesson will be to gain experience in solving more complex problems to simplify rational expressions.

Example 1.

Solution. At first, it may seem that the indicated fractions can be canceled, since the expressions in the numerators of the fractions are very similar to the formulas for the perfect squares of their corresponding denominators. In this case, it is important not to rush, but to separately check whether this is so.

Let's check the numerator of the first fraction:. Now the numerator is the second:.

As you can see, our expectations were not met, and the expressions in the numerators are not perfect squares, since they do not have a doubling of the product. Such expressions, if we recall the course of the 7th grade, are called incomplete squares. You should be very careful in such cases, since confusing the formula of a complete square with an incomplete square is a very common mistake, and such examples test the student's attentiveness.

Since the cancellation is impossible, let's add the fractions. The denominators do not have common factors, so they are simply multiplied to get the lowest common denominator, and the complementary factor for each fraction is the denominator of the other fraction.

Of course, further you can open the brackets and then give similar terms, however, in this case, you can get by with less effort and notice that in the numerator the first term is the formula for the sum of cubes, and the second is the difference between cubes. For convenience, let us recall these formulas in general form:

In our case, the expressions in the numerator collapse as follows:

, the second expression is the same. We have:

Answer..

Example 2. Simplify rational expression .

Solution. This example is similar to the previous one, but here you can immediately see that there are incomplete squares in the numerators of the fractions, so reduction at the initial stage of the solution is impossible. Similarly to the previous example, add the fractions:

Here, similarly to the method indicated above, we noticed and collapsed the expressions according to the formulas for the sum and difference of cubes.

Answer..

Example 3. Simplify rational expression.

Solution. You can see that the denominator of the second fraction is factorized using the formula for the sum of cubes. As we already know, factoring the denominators is useful for further finding the lowest common denominator of fractions.

We indicate the smallest common denominator of fractions, it is equal to:, because it is divided by the denominator of the third fraction, and the first expression is generally an integer, and any denominator is suitable for it. Having indicated the obvious additional factors, we write:

Answer.

Let's look at a more complex example with “multi-level” fractions.

Example 4. Prove the identity for all admissible values of the variable.

Proof. To prove the indicated identity, we will try to simplify its left-hand side (complex) to the simple form that is required of us. To do this, we will perform all actions with fractions in the numerator and denominator, and then divide the fractions and simplify the result.

Proven for all admissible values of the variable.

Proven.

In the next lesson, we will take a closer look at more complex examples for transforming rational expressions.

Bibliography

1. Bashmakov M.I. Algebra grade 8. - M .: Education, 2004.

2. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra 8. - 5th ed. - M .: Education, 2010.

3. Nikolsky S.M., Potapov M.A., Reshetnikov N.N., Shevkin A.V. Algebra grade 8. Textbook for educational institutions. - M .: Education, 2006.

2. Development of lessons, presentations, class notes ().

Homework

1. No. 96-101. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra 8. - 5th ed. - M .: Education, 2010.

2. Simplify the expression .

3. Simplify the expression.

4. Prove the identity.

Lesson and presentation on the topic: "Transformation of rational expressions. Examples of problem solving"

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The concept of rational expression

The concept of "rational expression" is similar to the concept of "rational fraction". The expression is also represented as a fraction. Only in the numerators we have - not numbers, but various kinds of expressions. Most often these are polynomials. An algebraic fraction is a fractional expression consisting of numbers and variables.

When solving many problems in elementary grades, after performing arithmetic operations, we received specific numerical values, most often fractions. Now, after performing the operations, we will receive algebraic fractions. Guys, remember: to get the right answer, you need to simplify the expression you are working with as much as possible. One should get the smallest degree possible; the same expressions in the numerators and denominators should be reduced; with expressions that can be collapsed, you must do so. That is, after performing a series of actions, we should get the most simple algebraic fraction.

Rational Expression Procedure

The procedure for performing operations with rational expressions is the same as for arithmetic operations. First, the actions in parentheses are performed, then multiplication and division, raising to a power, and finally addition and subtraction.

To prove an identity means to show that for all values of the variables the right and left sides are equal. There are many examples of proof of identities.

The main methods for solving identities are.

Convert left side to equal right side.
Convert right side to equal left side.
Transform the left and right sides separately until you get the same expression.
Subtract the right from the left side, and as a result, you should get zero.

Convert rational expressions. Examples of problem solving

Example 1.
Prove the identity:

$ (\ frac (a + 5) (5a-1) + \ frac (a + 5) (a + 1)): (\ frac (a ^ 2 + 5a) (1-5a)) + \ frac (a ^ 2 + 5) (a + 1) = a-1 $.

Solution.
Obviously, we need to transform the left side.
First, let's perform the actions in parentheses:

1) $ \ frac (a + 5) (5a-1) + \ frac (a + 5) (a + 1) = \ frac ((a + 5) (a + 1) + (a + 5) (5a -1)) ((a + 1) (5a-1)) = $
$ = \ frac ((a + 5) (a + 1 + 5a-1)) ((a + 1) (5a-1)) = \ frac ((a + 5) (6a)) ((a + 1 ) (5a-1)) $

You should try to take out common factors to the maximum.
2) We transform the expression by which we divide:

$ \ frac (a ^ 2 + 5a) (1-5a) = \ frac (a (a + 5)) ((1-5a) = \ frac (a (a + 5)) (- (5a-1) ) $

.
3) Let's perform the division operation:

$ \ frac ((a + 5) (6a)) ((a + 1) (5a-1)): \ frac (a (a + 5)) (- (5a-1)) = \ frac ((a +5) (6a)) ((a + 1) (5a-1)) * \ frac (- (5a-1)) (a (a + 5)) = \ frac (-6) (a + 1) $.

4) Let's perform the addition operation:

$ \ frac (-6) (a + 1) + \ frac (a ^ 2 + 5) (a + 1) = \ frac (a ^ 2-1) (a + 1) = \ frac ((a-1 ) (a + 1)) (a +)) = a-1 $.

The right and left sides have coincided. Hence, the identity is proved.
Guys, when solving this example, we needed knowledge of many formulas and operations. We see that after the transformation, the large expression turned into a very small one. When solving almost all problems, transformations usually lead to simple expressions.

Example 2.
Simplify the expression:

$ (\ frac (a ^ 2) (a + b) - \ frac (a ^ 3) (a ^ 2 + 2ab + b ^ 2)): (\ frac (a) (a + b) - \ frac ( a ^ 2) (a ^ 2-b ^ 2)) $.

Solution.
Let's start with the first parentheses.

1. $ \ frac (a ^ 2) (a + b) - \ frac (a ^ 3) (a ^ 2 + 2ab + b ^ 2) = \ frac (a ^ 2) (a + b) - \ frac (a ^ 3) ((a + b) ^ 2) = \ frac (a ^ 2 (a + b) -a ^ 3) ((a + b) ^ 2) = $
$ = \ frac (a ^ 3 + a ^ 2 b-a ^ 3) ((a + b) ^ 2) = \ frac (a ^ 2b) ((a + b) ^ 2) $.

2. Let's transform the second brackets.

$ \ frac (a) (a + b) - \ frac (a ^ 2) (a ^ 2-b ^ 2) = \ frac (a) (a + b) - \ frac (a ^ 2) ((ab ) (a + b)) = \ frac (a (ab) -a ^ 2) ((ab) (a + b)) = $
$ = \ frac (a ^ 2-ab-a ^ 2) ((a-b) (a + b)) = \ frac (-ab) ((a-b) (a + b)) $.

3. Let's do the division.

$ \ frac (a ^ 2b) ((a + b) ^ 2): \ frac (-ab) ((ab) (a + b)) = \ frac (a ^ 2b) ((a + b) ^ 2 ) * \ frac ((ab) (a + b)) ((- ab)) = $
$ = - \ frac (a (a-b)) (a + b) $

Answer: $ - \ frac (a (a-b)) (a + b) $.

Example 3.
Follow the steps:

$ \ frac (k-4) (k-2): (\ frac (80k) ((k ^ 3-8) + \ frac (2k) (k ^ 2 + 2k + 4) - \ frac (k-16 ) (2-k)) - \ frac (6k + 4) ((4-k) ^ 2) $.

Solution.
As always, you should start with parentheses.

1. $ \ frac (80k) (k ^ 3-8) + \ frac (2k) (k ^ 2 + 2k + 4) - \ frac (k-16) (2-k) = \ frac (80k) ( (k-2) (k ^ 2 + 2k + 4)) + \ frac (2k) (k ^ 2 + 2k + 4) + \ frac (k-16) (k-2) = $

$ = \ frac (80k + 2k (k-2) + (k-16) (k ^ 2 + 2k + 4)) ((k-2) (k ^ 2 + 2k + 4)) = \ frac (80k + 2k ^ 2-4k + k ^ 3 + 2k ^ 2 + 4k-16k ^ 2-32k-64) ((k-2) (k ^ 2 + 2k + 4)) = $

$ = \ frac (k ^ 3-12k ^ 2 + 48k-64) ((k-2) (k ^ 2 + 2k + 4)) = \ frac ((k-4) ^ 3) ((k-2 ) (k ^ 2 + 2k + 4)) $.

2. Now let's do the division.

$ \ frac (k-4) (k-2): \ frac ((k-4) ^ 3) ((k-2) (k ^ 2 + 2k + 4)) = \ frac (k-4) ( k-2) * \ frac ((k-2) (k ^ 2 + 2k + 4)) ((k-4) ^ 3) = \ frac ((k ^ 2 + 2k + 4)) ((k- 4) ^ 2) $.

3. Let's use the property: $ (4-k) ^ 2 = (k-4) ^ 2 $.
4. Let's perform the subtraction operation.

$ \ frac ((k ^ 2 + 2k + 4)) ((k-4) ^ 2) - \ frac (6k + 4) ((k-4) ^ 2) = \ frac (k ^ 2-4k) ((k-4) ^ 2) = \ frac (k (k-4)) ((k-4) ^ 2) = \ frac (k) (k-4) $.

As we said earlier, you need to simplify the fraction as much as possible.
Answer: $ \ frac (k) (k-4) $.

Tasks for independent solution

1. Prove the identity:

$ \ frac (b ^ 2-14) (b-4) - (\ frac (3-b) (7b-4) + \ frac (b-3) (b-4)) * \ frac (4-7b ) (9b-3b ^ 2) = b + 4 $.

2. Simplify the expression:

$ \ frac (4 (z + 4) ^ 2) (z-2) * (\ frac (z) (2z-4) - \ frac (z ^ 2 + 4) (2z ^ 2-8) - \ frac (2) (z ^ 2 + 2z)) $.

3. Follow the steps:

$ (\ frac (ab) (a ^ 2 + 2ab + b ^ 2) - \ frac (2a) ((ab) (a + b)) + \ frac (ab) ((ab) ^ 2)) * \ frac (a ^ 4-b ^ 4) (8ab ^ 2) + \ frac (2b ^ 2) (a ^ 2-b ^ 2) $.

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

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

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

To fix the solution yourself, take a few examples:

Examples:

Solutions:

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

The first step is to factorize:

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

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

Let's remember:

Answers:

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

2. Here the common denominator is:

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

It is 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;

· We determine common (identical) factors;

· Write out all common factors one time;

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

So, in order:

1) we decompose the denominators into factors:

2) we determine the common (identical) factors:

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

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

By the way, there is one trick:

For example: .

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

to the extent

in degree.

Let's complicate the task:

How do you make fractions the same denominator?

Let's remember the basic property of a fraction:

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

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

So, another unshakable rule:

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

But what must be multiplied by in order to receive?

Here on and multiply. And multiply by:

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

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

What do you think about expression? Is it elementary?

No, since it can be factorized:

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

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

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

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

Another example:

Solution:

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

Fine! Then:

Another example:

Solution:

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

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

So we will write:

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

Now we bring to a common denominator:

Got it? Let's check it out now.

Tasks for independent solution:

Answers:

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

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

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

What if there are already three fractions?

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

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

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

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

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

Exactly what is needed!

5. Multiplication and division of fractions.

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

Procedure

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

Did you count it?

It should work.

So, let me remind you.

The first step is to calculate the degree.

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

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

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

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

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

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

Okay, it's all simple.

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

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

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

For 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:

So that is all. Nothing complicated, right?

Another example:

Simplify the expression.

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

Solution:

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

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

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

I will schematically number the steps:

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

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

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

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

And promised at the very beginning:

Answers:

Solutions (concise):

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

Now forward to learning!

TRANSFORMATION OF EXPRESSIONS. SUMMARY AND BASIC FORMULAS

Basic simplification operations:

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