Converting fractional expressions containing the action of multiplication. Converting Expressions. Detailed Theory (2019)

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

Subject: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 operations of addition, subtraction, multiplication and division (raising to a power).

By the concept of “transforming 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), dividing numbers, and then adding/subtracting operations.

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

Example 1.

Solution. At first it may seem that these fractions can be reduced, since the expressions in the numerators of 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 second numerator: .

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

Since reduction is impossible, we will perform the addition of fractions. The denominators do not have common factors, so they are simply multiplied to obtain the lowest common denominator, and the additional factor for each fraction is the denominator of the other fraction.

Of course, you can then open the brackets and then bring 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 of cubes. For convenience, let us recall these formulas in general form:

In our case, the expressions in the numerator are collapsed as follows:

, the second expression is similar. We have:

Answer..

Example 2. Simplify rational expression .

Solution. This example is similar to the previous one, but here it is immediately clear that the numerators of the fractions contain partial squares, so the reduction by initial stage solutions are impossible. Similarly to the previous example, we add the fractions:

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

Answer..

Example 3. Simplify a rational expression.

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

Let us indicate the lowest common denominator of the fractions, it is equal to: , since 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 consider a more complex example with “multi-story” fractions.

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

Proof. To prove this identity, we will try to simplify it left side(complicated) before simple type which is required of us. To do this, we will perform all the operations with fractions in the numerator and denominator, and then divide the fractions and simplify the result.

Proven for all permissible values of the variable.

Proven.

In the next lesson we will take a closer look at more complex examples to transform rational expressions.

Bibliography

1. Bashmakov M.I. Algebra 8th grade. - 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 8th grade. Tutorial for educational institutions. - M.: Education, 2006.

2. Lesson developments, presentations, lesson 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.

>>Math: Converting rational expressions

Converting rational expressions

This paragraph summarizes everything that we, starting from the 7th grade, talked about mathematical language, mathematical symbolism, numbers, variables, powers, polynomials and algebraic fractions. But first, let's take a short excursion into the past.

Remember how things were in the elementary grades with the study of numbers and numerical expressions.

And, say, only one label can be attached to a fraction - a rational number.

The situation is similar with algebraic expressions: the first stage of their study is numbers, variables, degrees (“digits”); the second stage of their study is monomials (“natural numbers”); the third stage of their study is polynomials (“integers”); the fourth stage of their study - algebraic fractions
(“rational numbers”). Moreover, each next stage, as it were, absorbs the previous one: for example, numbers, variables, powers are special cases of monomials; monomials - special cases of polynomials; polynomials - special cases algebraic fractions. By the way, the following terms are sometimes used in algebra: polynomial - integer expression, an algebraic fraction is a fractional expression (this only strengthens the analogy).

Let's continue the above analogy. You know that any numerical expression, after performing all the arithmetic operations included in its composition, takes on a specific numerical value - a rational number (of course, it can also turn out to be natural number, both a whole number and a fraction - it doesn’t matter). Similarly, any algebraic expression composed of numbers and variables using arithmetic operations and raising to natural numbers degree, after performing the transformations takes the form of an algebraic fraction and again, in particular, the result may not be a fraction, but a polynomial or even a monomial). For such expressions in algebra the term rational expression is used.

Example. Prove identity

Solution.
To prove an identity means to establish that for all permissible values of the variables, its left and right sides are identically equal expressions. In algebra, identities are proven different ways:

1) perform transformations on the left side and ultimately obtain the right side;

2) perform transformations on the right side and ultimately obtain the left side;

3) transform the right and left sides separately and obtain the same expression in both the first and second cases;

4) make up the difference between the left and right parts and as a result of its transformations they get zero.

Which method to choose depends on the specific type identities which you are asked to prove. In this example, it is advisable to choose the first method.

To convert rational expressions, the same procedure is adopted as for converting numerical expressions. This means that first they perform the actions in brackets, then the actions of the second stage (multiplication, division, exponentiation), then the actions of the first stage (addition, subtraction).

Let's carry out transformations based on the rules algorithms that were developed in the previous paragraphs.

As you can see, we were able to transform the left side of the identity being verified to the form of the right side. This means that the identity is proven. However, recall that the identity is valid only for admissible values of the variables. In this example, these are any values of a and b, except those that make the denominators of the fractions zero. This means that any pairs of numbers (a; b) are valid, except those for which at least one of the equalities is satisfied:

2a - b = 0, 2a + b = 0, b = 0.

Mordkovich A. G., Algebra. 8th grade: Textbook. for general education institutions. - 3rd ed., revised. - M.: Mnemosyne, 2001. - 223 p.: ill.

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The article talks about the transformation of rational expressions. Let's consider the types of rational expressions, their transformations, groupings, and bracketing the common factor. Let's learn to represent fractional rational expressions in the form of rational fractions.

Yandex.RTB R-A-339285-1

Definition and examples of rational expressions

Definition 1

Expressions that are made up of numbers, variables, parentheses, powers with the operations of addition, subtraction, multiplication, division with the presence of a fraction line are called rational expressions.

For example, we have that 5, 2 3 x - 5, - 3 a b 3 - 1 c 2 + 4 a 2 + b 2 1 + a: (1 - b) , (x + 1) (y - 2) x 5 - 5 · x · y · 2 - 1 11 · x 3 .

That is, these are expressions that are not divided into expressions with variables. The study of rational expressions begins in grade 8, where they are called fractional rational expressions. Particular attention is paid to fractions in the numerator, which are transformed using transformation rules.

This allows us to proceed to the transformation of rational fractions of arbitrary form. Such an expression can be considered as an expression with the presence of rational fractions and integer expressions with action signs.

Main types of transformations of rational expressions

Rational expressions are used to perform identical transformations, groupings, bringing similar ones, and performing other operations with numbers. The purpose of such expressions is simplification.

Example 1

Convert the rational expression 3 · x x · y - 1 - 2 · x x · y - 1 .

Solution

It can be seen that such a rational expression is the difference between 3 x x y - 1 and 2 x x y - 1. We notice that their denominator is identical. This means that the reduction of similar terms will take the form

3 x x y - 1 - 2 x x y - 1 = x x y - 1 3 - 2 = x x y - 1

Answer: 3 · x x · y - 1 - 2 · x x · y - 1 = x x · y - 1 .

Example 2

Convert 2 x y 4 (- 4) x 2: (3 x - x) .

Solution

Initially, we perform the actions in brackets 3 · x − x = 2 · x. We present this expression in the form 2 · x · y 4 · (- 4) · x 2: (3 · x - x) = 2 · x · y 4 · (- 4) · x 2: 2 · x. We arrive at an expression that contains operations with one step, that is, it has addition and subtraction.

We get rid of parentheses by using the division property. Then we get that 2 · x · y 4 · (- 4) · x 2: 2 · x = 2 · x · y 4 · (- 4) · x 2: 2: x.

We group numerical factors with the variable x, after which we can perform operations with powers. We get that

2 x y 4 (- 4) x 2: 2: x = (2 (- 4) : 2) (x x 2: x) y 4 = - 4 x 2 y 4

Answer: 2 x y 4 (- 4) x 2: (3 x - x) = - 4 x 2 y 4.

Example 3

Transform an expression of the form x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 .

Solution

First, we transform the numerator and denominator. Then we get an expression of the form (x · (x + 3) - (3 · x + 1)): 1 2 · x · 4 + 2, and the actions in parentheses are done first. In the numerator, operations are performed and factors are grouped. Then we get an expression of the form x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 = x 2 + 3 · x - 3 · x - 1 1 2 · 4 · x + 2 = x 2 - 1 2 · x + 2 .

We transform the difference of squares formula in the numerator, then we get that

x 2 - 1 2 x + 2 = (x - 1) (x + 1) 2 (x + 1) = x - 1 2

Answer: x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 = x - 1 2 .

Rational fraction representation

Algebraic fractions are most often simplified when solved. Every rational is reduced to this different ways. Everything needs to be done necessary actions with polynomials so that a rational expression can ultimately give a rational fraction.

Example 4

Present as a rational fraction a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a.

Solution

This expression can be represented as a 2 - 25 a + 3 · 1 a 2 + 5 · a. Multiplication is performed primarily according to the rules.

We should start with multiplication, then we get that

a 2 - 25 a + 3 1 a 2 + 5 a = a - 5 (a + 5) a + 3 1 a (a + 5) = a - 5 (a + 5) 1 ( a + 3) a (a + 5) = a - 5 (a + 3) a

We present the obtained result with the original one. We get that

a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a = a + 5 a · a - 3 - a - 5 a + 3 · a

Now let's do the subtraction:

a + 5 a · a - 3 - a - 5 a + 3 · a = a + 5 · a + 3 a · (a - 3) · (a + 3) - (a - 5) · (a - 3) (a + 3) a (a - 3) = = a + 5 a + 3 - (a - 5) (a - 3) a (a - 3) (a + 3) = a 2 + 3 a + 5 a + 15 - (a 2 - 3 a - 5 a + 15) a (a - 3) (a + 3) = = 16 a a (a - 3) (a + 3) = 16 a - 3 (a + 3) = 16 a 2 - 9

After which it is obvious that the original expression will take the form 16 a 2 - 9.

Answer: a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a = 16 a 2 - 9 .

Example 5

Express x x + 1 + 1 2 · x - 1 1 + x as a rational fraction.

Solution

The given expression is written as a fraction, the numerator of which has x x + 1 + 1, and the denominator 2 x - 1 1 + x. It is necessary to make transformations x x + 1 + 1 . To do this you need to add a fraction and a number. We get that x x + 1 + 1 = x x + 1 + 1 1 = x x + 1 + 1 · (x + 1) 1 · (x + 1) = x x + 1 + x + 1 x + 1 = x + x + 1 x + 1 = 2 x + 1 x + 1

It follows that x x + 1 + 1 2 x - 1 1 + x = 2 x + 1 x + 1 2 x - 1 1 + x

The resulting fraction can be written as 2 x + 1 x + 1: 2 x - 1 1 + x.

After division we arrive at a rational fraction of the form

2 x + 1 x + 1: 2 x - 1 1 + x = 2 x + 1 x + 1 1 + x 2 x - 1 = 2 x + 1 (1 + x) (x + 1) (2 x - 1) = 2 x + 1 2 x - 1

You can solve this differently.

Instead of dividing by 2 x - 1 1 + x, we multiply by its inverse 1 + x 2 x - 1. Let us apply the distribution property and find that

x x + 1 + 1 2 x - 1 1 + x = x x + 1 + 1: 2 x - 1 1 + x = x x + 1 + 1 1 + x 2 x - 1 = = x x + 1 1 + x 2 x - 1 + 1 1 + x 2 x - 1 = x 1 + x (x + 1) 2 x - 1 + 1 + x 2 x - 1 = = x 2 x - 1 + 1 + x 2 x - 1 = x + 1 + x 2 x - 1 = 2 x + 1 2 x - 1

Answer: x x + 1 + 1 2 · x - 1 1 + x = 2 · x + 1 2 · x - 1 .

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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 our numerators are 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 the elementary grades, after performing arithmetic operations, we received specific numerical values, most often fractions. Now after performing the operations we will obtain algebraic fractions. Guys, remember: to get the correct answer, you need to simplify the expression you are working with as much as possible. One must obtain the smallest degree possible; identical expressions in 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 obtain the simplest possible algebraic fraction.

Procedure with rational expressions

The procedure for performing operations with rational expressions is the same as for arithmetic operations. First, the operations in parentheses are performed, then multiplication and division, exponentiation, 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 a lot of examples of proving identities.

The main ways to solve identities include.

Transform the left side to be equal to the right side.
Transform the right side to be equal to the left.
Transform the left and right sides separately until you get the same expression.
The right side is subtracted from the left side, and the result should be zero.

Converting 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 do the steps 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 apply common factors to the maximum.
2) 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) 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) 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 parts coincided. This means the identity is proven.
Guys, when solving this example we needed knowledge of many formulas and operations. We see that after the transformation, the large expression has 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 brackets.

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. Transform the second brackets.

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

3. Let's do the division.

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

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

Example 3.
Follow these 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 need to start with the brackets.

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: $\frac(k)(k-4)$.

Problems to solve independently

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 these steps:

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

From the school algebra course we move on to specifics. In this article we will study in detail a special type of rational expressions - rational fractions, and also consider what characteristic identical conversions of rational fractions take place.

Let us immediately note that rational fractions in the sense in which we define them below are called algebraic fractions in some algebra textbooks. That is, in this article we will understand rational and algebraic fractions as the same thing.

As usual, let's start with a definition and examples. Next we’ll talk about bringing a rational fraction to a new denominator and changing the signs of the members of the fraction. After this, we will look at how to reduce fractions. Finally, let's look at representing a rational fraction as a sum of several fractions. We will provide all information with examples detailed descriptions decisions.

Page navigation.

Definition and examples of rational fractions

Rational fractions are studied in 8th grade algebra lessons. We will use the definition of a rational fraction, which is given in the algebra textbook for 8th grade by Yu. N. Makarychev et al.

This definition does not specify whether the polynomials in the numerator and denominator of a rational fraction must be polynomials of the standard form or not. Therefore, we will assume that the notations for rational fractions can contain both standard and non-standard polynomials.

Here are a few examples of rational fractions. So, x/8 and - rational fractions. And fractions and do not fit the stated definition of a rational fraction, since in the first of them the numerator does not contain a polynomial, and in the second, both the numerator and the denominator contain expressions that are not polynomials.

Converting the numerator and denominator of a rational fraction

The numerator and denominator of any fraction are self-sufficient mathematical expressions, in the case of rational fractions, these are polynomials; in a particular case, monomials and numbers. Therefore, identical transformations can be carried out with the numerator and denominator of a rational fraction, as with any expression. In other words, the expression in the numerator of a rational fraction can be replaced by an identically equal expression, just like the denominator.

You can perform identical transformations in the numerator and denominator of a rational fraction. For example, in the numerator you can group and reduce similar terms, and in the denominator you can replace the product of several numbers with its value. And since the numerator and denominator of a rational fraction are polynomials, it is possible to perform transformations characteristic of polynomials with them, for example, reduction to a standard form or representation in the form of a product.

For clarity, let's consider solutions to several examples.

Example.

Convert rational fraction so that the numerator contains a polynomial of standard form, and the denominator contains the product of polynomials.

Solution.

Reducing rational fractions to a new denominator is mainly used in adding and subtracting rational fractions.

Changing signs in front of a fraction, as well as in its numerator and denominator

The main property of a fraction can be used to change the signs of the members of a fraction. Indeed, multiplying the numerator and denominator of a rational fraction by -1 is equivalent to changing their signs, and the result is a fraction identically equal to the given one. This transformation has to be used quite often when working with rational fractions.

Thus, if you simultaneously change the signs of the numerator and denominator of a fraction, you will get a fraction equal to the original one. This statement is answered by equality.

Let's give an example. A rational fraction can be replaced by an identically equal fraction with changed signs of the numerator and denominator of the form.

With fractions, you can carry out another identical transformation, in which the sign of either the numerator or the denominator changes. Let us state the corresponding rule. If you replace the sign of a fraction together with the sign of the numerator or denominator, you get a fraction that is identically equal to the original one. The written statement corresponds to the equalities and .

Proving these equalities is not difficult. The proof is based on the properties of multiplication of numbers. Let's prove the first of them: . Using similar transformations, the equality is proved.

For example, a fraction can be replaced by the expression or.

To conclude this point, we present two more useful equalities and . That is, if you change the sign of only the numerator or only the denominator, the fraction will change its sign. For example, And .

The considered transformations, which allow changing the sign of the terms of a fraction, are often used when transforming fractional rational expressions.

Reducing rational fractions

The following transformation of rational fractions, called reduction of rational fractions, is based on the same basic property of a fraction. This transformation corresponds to the equality , where a, b and c are some polynomials, and b and c are non-zero.

From the above equality it becomes clear that reducing a rational fraction implies getting rid of the common factor in its numerator and denominator.

Example.

Cancel a rational fraction.

Solution.

The common factor 2 is immediately visible, let’s perform a reduction by it (when writing, it is convenient to cross out the common factors that are being reduced by). We have . Since x 2 =x x and y 7 =y 3 y 4 (see if necessary), it is clear that x is a common factor of the numerator and denominator of the resulting fraction, as is y 3. Let's reduce by these factors: . This completes the reduction.

Above we carried out the reduction of rational fractions sequentially. Or it was possible to perform the reduction in one step, immediately reducing the fraction by 2 x y 3. In this case, the solution would look like this: .

Answer:

When reducing rational fractions, the main problem is that the common factor of the numerator and denominator is not always visible. Moreover, it does not always exist. In order to find a common factor or verify its absence, you need to factor the numerator and denominator of a rational fraction. If there is no common factor, then the original rational fraction does not need to be reduced, otherwise, reduction is carried out.

Various nuances can arise in the process of reducing rational fractions. The main subtleties are discussed in the article reducing algebraic fractions using examples and in detail.

Concluding the conversation about the reduction of rational fractions, we note that this transformation is identical, and the main difficulty in its implementation lies in factoring the polynomials in the numerator and denominator.

Representation of a rational fraction as a sum of fractions

Quite specific, but in some cases very useful, is the transformation of a rational fraction, which consists in its representation as the sum of several fractions, or the sum of an entire expression and a fraction.

A rational fraction, the numerator of which contains a polynomial representing the sum of several monomials, can always be written as a sum of fractions with same denominators, the numerators of which contain the corresponding monomials. For example, . This representation is explained by the rule for adding and subtracting algebraic fractions with like denominators.

In general, any rational fraction can be represented as a sum of fractions in many different ways. For example, the fraction a/b can be represented as the sum of two fractions - an arbitrary fraction c/d and a fraction equal to the difference between the fractions a/b and c/d. This statement is true, since the equality holds . For example, a rational fraction can be represented as a sum of fractions in various ways: Let's imagine the original fraction as the sum of an integer expression and a fraction. By dividing the numerator by the denominator with a column, we get the equality . The value of the expression n 3 +4 for any integer n is an integer. And the value of a fraction is an integer if and only if its denominator is 1, −1, 3, or −3. These values correspond to the values n=3, n=1, n=5 and n=−1, respectively.

Answer:

−1 , 1 , 3 , 5 .

Bibliography.

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