Operations with algebraic fractions. Basic Concepts Represent as an algebraic fraction

This lesson covers the concept of an algebraic fraction. People encounter fractions in the simplest life situations: when it is necessary to divide an object into several parts, for example, to cut a cake equally into ten people. Obviously, everyone gets a piece of the cake. In this case, we are faced with the concept of a numerical fraction, but a situation is possible when an object is divided into an unknown number of parts, for example, by x. In this case, the concept of a fractional expression arises. You have already become acquainted with whole expressions (not containing division into expressions with variables) and their properties in 7th grade. Next we will look at the concept of a rational fraction, as well as acceptable values of variables.

Rational expressions are divided into integer and fractional expressions.

Definition.Rational fraction is a fractional expression of the form , where are polynomials. - numerator, - denominator.

Examplesrational expressions:- fractional expressions; - whole expressions. In the first expression, for example, the numerator is , and the denominator is .

Meaning algebraic fraction like anyone algebraic expression, depends on the numerical value of the variables that are included in it. In particular, in the first example the value of the fraction depends on the values of the variables and , and in the second example only on the value of the variable .

Let's consider the first typical task: calculating the value rational fraction for different values of the variables included in it.

Example 1. Calculate the value of the fraction for a) , b) , c)

Solution. Let's substitute the values of the variables into the indicated fraction: a) , b) , c) - does not exist (since you cannot divide by zero).

Answer: a) 3; b) 1; c) does not exist.

As you can see, two typical problems arise for any fraction: 1) calculating the fraction, 2) finding valid and invalid values letter variables.

Definition.Valid Variable Values- values of variables at which the expression makes sense. The set of all possible values of variables is called ODZ or domain of definition.

The value of literal variables may be invalid if the denominator of the fraction at these values is zero. In all other cases, the values of the variables are valid, since the fraction can be calculated.

Example 2.

Solution. For this expression to make sense, it is necessary and sufficient that the denominator of the fraction does not equal zero. Thus, only those values of the variable will be invalid for which the denominator is equal to zero. The denominator of the fraction is , so we solve the linear equation:

Therefore, given the value of the variable, the fraction has no meaning.

Answer: -5.

From the solution of the example, the rule for finding invalid values of variables follows - the denominator of the fraction is equal to zero and the roots of the corresponding equation are found.

Let's look at several similar examples.

Example 3. Establish at what values of the variable the fraction does not make sense .

Solution..

Answer..

Example 4. Establish at what values of the variable the fraction does not make sense.

Solution..

There are other formulations of this problem - find domain of definition or range of acceptable expression values (APV). This means finding all valid values of the variables. In our example, these are all values except . It is convenient to depict the domain of definition on a number axis.

To do this, we will cut out a point on it, as indicated in the figure:

Rice. 1

Thus, fraction definition domain there will be all numbers except 3.

Answer..

Example 5. Establish at what values of the variable the fraction does not make sense.

Solution..

Let us depict the resulting solution on the numerical axis:

Rice. 2

Answer..

Example 6.

Solution.. We have obtained the equality of two variables, we will give numerical examples: or, etc.

Let us depict this solution on a graph in the Cartesian coordinate system:

Rice. 3. Graph of a function

The coordinates of any point lying on this graph are not included in the range of acceptable fraction values.

Answer..

In the examples discussed, we encountered a situation where division by zero occurred. Now consider the case where a more interesting situation arises with type division.

Example 7. Establish at what values of the variables the fraction does not make sense.

Solution..

It turns out that the fraction makes no sense at . But one could argue that this is not the case because: .

It may seem that if the final expression is equal to 8 at , then the original one can also be calculated, and therefore makes sense at . However, if we substitute it into the original expression, we get - it makes no sense.

Answer..

To understand this example in more detail, let’s solve the following problem: at what values does the indicated fraction equal zero?

This article continues the topic of converting algebraic fractions: consider such an action as reducing algebraic fractions. Let's define the term itself, formulate a reduction rule and analyze practical examples.

The meaning of reducing an algebraic fraction

In materials about common fractions, we looked at its reduction. We defined reducing a fraction as dividing its numerator and denominator by a common factor.

Reducing an algebraic fraction is a similar operation.

Definition 1

Reducing an algebraic fraction is the division of its numerator and denominator by a common factor. In this case, in contrast to the reduction of an ordinary fraction (the common denominator can only be a number), the common factor of the numerator and denominator of an algebraic fraction can be a polynomial, in particular, a monomial or a number.

For example, the algebraic fraction 3 x 2 + 6 x y 6 x 3 y + 12 x 2 y 2 can be reduced by the number 3, resulting in: x 2 + 2 x y 6 x 3 · y + 12 · x 2 · y 2 . We can reduce the same fraction by the variable x, and this will give us the expression 3 x + 6 y 6 x 2 y + 12 x y 2. It is also possible to reduce a given fraction by a monomial 3 x or any of the polynomials x + 2 y, 3 x + 6 y , x 2 + 2 x y or 3 x 2 + 6 x y.

The ultimate goal of reducing an algebraic fraction is a fraction of a simpler form, at best an irreducible fraction.

Are all algebraic fractions subject to reduction?

Again, from materials on ordinary fractions, we know that there are reducible and irreducible fractions. Irreducible fractions are fractions that do not have common numerator and denominator factors other than 1.

It’s the same with algebraic fractions: they may have common factors in the numerator and denominator, or they may not. The presence of common factors allows you to simplify the original fraction through reduction. When there are no common factors, it is impossible to optimize a given fraction using the reduction method.

In general cases, given the type of fraction it is quite difficult to understand whether it can be reduced. Of course, in some cases the presence of a common factor between the numerator and denominator is obvious. For example, in the algebraic fraction 3 x 2 3 y it is quite clear that the common factor is the number 3.

In the fraction - x · y 5 · x · y · z 3 we also immediately understand that it can be reduced by x, or y, or x · y. And yet, much more often there are examples of algebraic fractions, when the common factor of the numerator and denominator is not so easy to see, and even more often, it is simply absent.

For example, we can reduce the fraction x 3 - 1 x 2 - 1 by x - 1, while the specified common factor is not present in the entry. But the fraction x 3 - x 2 + x - 1 x 3 + x 2 + 4 · x + 4 cannot be reduced, since the numerator and denominator do not have a common factor.

Thus, the question of determining the reducibility of an algebraic fraction is not so simple, and it is often easier to work with a fraction of a given form than to try to find out whether it is reducible. In this case, such transformations take place that in particular cases make it possible to determine the common factor of the numerator and denominator or to draw a conclusion about the irreducibility of a fraction. We will examine this issue in detail in the next paragraph of the article.

Rule for reducing algebraic fractions

Rule for reducing algebraic fractions consists of two sequential actions:

finding common factors of the numerator and denominator;
if any are found, the action of reducing the fraction is carried out directly.

The most convenient method of finding common denominators is to factor the polynomials present in the numerator and denominator of a given algebraic fraction. This allows you to immediately clearly see the presence or absence of common factors.

The very action of reducing an algebraic fraction is based on the main property of an algebraic fraction, expressed by the equality undefined, where a, b, c are some polynomials, and b and c are non-zero. The first step is to reduce the fraction to the form a · c b · c, in which we immediately notice the common factor c. The second step is to perform a reduction, i.e. transition to a fraction of the form a b .

Typical examples

Despite some obviousness, let us clarify the special case when the numerator and denominator of an algebraic fraction are equal. Similar fractions are identically equal to 1 on the entire ODZ of the variables of this fraction:

5 5 = 1 ; - 2 3 - 2 3 = 1 ; x x = 1 ; - 3, 2 x 3 - 3, 2 x 3 = 1; 1 2 · x - x 2 · y 1 2 · x - x 2 · y ;

Since ordinary fractions are a special case of algebraic fractions, let us recall how they are reduced. The natural numbers written in the numerator and denominator are factored into prime factors, then the common factors are canceled (if any).

For example, 24 1260 = 2 2 2 3 2 2 3 3 5 7 = 2 3 5 7 = 2 105

The product of simple identical factors can be written as powers, and in the process of reducing a fraction, use the property of dividing powers with identical bases. Then the above solution would be:

24 1260 = 2 3 3 2 2 3 2 5 7 = 2 3 - 2 3 2 - 1 5 7 = 2 105

(numerator and denominator divided by a common factor 2 2 3). Or for clarity, based on the properties of multiplication and division, we give the solution the following form:

24 1260 = 2 3 3 2 2 3 2 5 7 = 2 3 2 2 3 3 2 1 5 7 = 2 1 1 3 1 35 = 2 105

By analogy, the reduction of algebraic fractions is carried out, in which the numerator and denominator have monomials with integer coefficients.

Example 1

The algebraic fraction is given - 27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z. It needs to be reduced.

Solution

It is possible to write the numerator and denominator of a given fraction as a product of simple factors and variables, and then carry out the reduction:

27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z = - 3 · 3 · 3 · a · a · a · a · a · b · b · c · z 2 · 3 · a · a · b · b · c · c · c · c · c · c · c · z = = - 3 · 3 · a · a · a 2 · c · c · c · c · c · c = - 9 a 3 2 c 6

However, a more rational way would be to write the solution as an expression with powers:

27 · a 5 · b 2 · c · z 6 · a 2 · b 2 · c 7 · z = - 3 3 · a 5 · b 2 · c · z 2 · 3 · a 2 · b 2 · c 7 · z = - 3 3 2 · 3 · a 5 a 2 · b 2 b 2 · c c 7 · z z = = - 3 3 - 1 2 · a 5 - 2 1 · 1 · 1 c 7 - 1 · 1 = · - 3 2 · a 3 2 · c 6 = · - 9 · a 3 2 · c 6 .

Answer:- 27 a 5 b 2 c z 6 a 2 b 2 c 7 z = - 9 a 3 2 c 6

When the numerator and denominator of an algebraic fraction contain fractional numerical coefficients, there are two possible ways of further action: either divide these fractional coefficients separately, or first get rid of the fractional coefficients by multiplying the numerator and denominator by some natural number. The last transformation is carried out due to the basic property of an algebraic fraction (you can read about it in the article “Reducing an algebraic fraction to a new denominator”).

Example 2

The given fraction is 2 5 x 0, 3 x 3. It needs to be reduced.

Solution

It is possible to reduce the fraction this way:

2 5 x 0, 3 x 3 = 2 5 3 10 x x 3 = 4 3 1 x 2 = 4 3 x 2

Let's try to solve the problem differently, having first gotten rid of fractional coefficients - multiply the numerator and denominator by the least common multiple of the denominators of these coefficients, i.e. on LCM (5, 10) = 10. Then we get:

2 5 x 0, 3 x 3 = 10 2 5 x 10 0, 3 x 3 = 4 x 3 x 3 = 4 3 x 2.

Answer: 2 5 x 0, 3 x 3 = 4 3 x 2

When we reduce general algebraic fractions, in which the numerators and denominators can be either monomials or polynomials, there can be a problem where the common factor is not always immediately visible. Or moreover, it simply does not exist. Then, to determine the common factor or record the fact of its absence, the numerator and denominator of the algebraic fraction are factored.

Example 3

The rational fraction 2 · a 2 · b 2 + 28 · a · b 2 + 98 · b 2 a 2 · b 3 - 49 · b 3 is given. It needs to be reduced.

Solution

Let us factor the polynomials in the numerator and denominator. Let's put it out of brackets:

2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49)

We see that the expression in parentheses can be converted using abbreviated multiplication formulas:

2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49) = 2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7)

It is clearly seen that it is possible to reduce a fraction by a common factor b 2 (a + 7). Let's make a reduction:

2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7) = 2 (a + 7) b (a - 7) = 2 a + 14 a b - 7 b

Let us write a short solution without explanation as a chain of equalities:

2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 b 2 (a 2 + 14 a + 49) b 3 (a 2 - 49) = = 2 b 2 (a + 7) 2 b 3 (a - 7) (a + 7) = 2 (a + 7) b (a - 7) = 2 a + 14 a b - 7 b

Answer: 2 a 2 b 2 + 28 a b 2 + 98 b 2 a 2 b 3 - 49 b 3 = 2 a + 14 a b - 7 b.

It happens that common factors are hidden by numerical coefficients. Then, when reducing fractions, it is optimal to put the numerical factors at higher powers of the numerator and denominator out of brackets.

Example 4

Given the algebraic fraction 1 5 · x - 2 7 · x 3 · y 5 · x 2 · y - 3 1 2 . It is necessary to reduce it if possible.

Solution

At first glance, the numerator and denominator do not have a common denominator. However, let's try to convert the given fraction. Let's take out the factor x in the numerator:

1 5 x - 2 7 x 3 y 5 x 2 y - 3 1 2 = x 1 5 - 2 7 x 2 y 5 x 2 y - 3 1 2

Now you can see some similarity between the expression in brackets and the expression in the denominator due to x 2 y . Let us take out the numerical coefficients of the higher powers of these polynomials:

x 1 5 - 2 7 x 2 y 5 x 2 y - 3 1 2 = x - 2 7 - 7 2 1 5 + x 2 y 5 x 2 y - 1 5 3 1 2 = = - 2 7 x - 7 10 + x 2 y 5 x 2 y - 7 10

Now the common factor becomes visible, we carry out the reduction:

2 7 x - 7 10 + x 2 y 5 x 2 y - 7 10 = - 2 7 x 5 = - 2 35 x

Answer: 1 5 x - 2 7 x 3 y 5 x 2 y - 3 1 2 = - 2 35 x .

Let us emphasize that the skill of reducing rational fractions depends on the ability to factor polynomials.

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Having received the initial information about fractions, we will move on to operations with algebraic fractions. You can perform any actions with them, including raising them to a power. When we do them, we end up with an algebraic fraction. All points must be analyzed sequentially.

Operations with algebraic fractions are similar to operations with ordinary fractions. Therefore, it is worth noting that the rules are the same for any actions performed with them.

Adding algebraic fractions

Addition can be performed in two cases: with the same denominators, with different denominators.

If you need to add fractions with the same denominators, you need to add the numerators and leave the denominator unchanged. This rule allows you to use the addition of fractions and polynomials that are in the numerators. We get that

a 2 + a b a b - 5 + 2 a b + 3 a b - 5 + 2 b 4 - 4 a b - 5 = a 2 + a b + 2 a b + 3 + 2 b 4 - 4 a b - 5 = = a 2 + 3 a b - 1 + 2 b 4 a b - 5

If there are numerators of a fraction with different numerators, then it is necessary to apply the rule: use reduction to a common denominator and add the resulting fractions.

Example 1

You need to add the fractions x x 2 - 1 and 3 x 2 - x

Solution

We reduce to a common denominator of the form x 2 x · x - 1 · x + 1 and 3 · x + 3 x · (x - 1) · (x + 1) .

Let's do the addition and get that

x 2 x (x - 1) (x + 1) + 3 x + 3 x (x - 1) (x + 1) = x 2 + 3 x + 3 x (x - 1) · (x + 1) = x 2 + 3 · x + 3 x 3 - x

Answer: x 2 + 3 x + 3 x 3 - x

The article on adding and subtracting such fractions has detailed information, which describes in detail each action performed on fractions. When performing addition, a reducible fraction may appear.

Subtraction

Subtraction is performed in the same way as addition. If the denominators are the same, the actions are performed only in the numerator, the denominator remains unchanged. For different denominators, a reduction to a common denominator is performed. Only after this can you begin to calculate.

Example 2

Let's move on to subtracting the fractions a + 5 a 2 + 2 and 1 - 2 · a 2 + a a 2 + 2.

Solution

It can be seen that the denominators are identical, which means a + 5 a 2 + 2 - 1 - 2 a 2 + a a 2 + 2 = a + 5 - (1 - 2 a 2 + a) a 2 + 2 = 2 a 2 + 4 a 2 + 2 .

Let's reduce the fraction 2 · a 2 + 4 a 2 + 2 = 2 · a 2 + 2 a 2 + 2 = 2.

Answer: 2

Example 3

Let's subtract 4 5 · x and 3 x - 1 .

Solution

The denominators are different, so we reduce it to the common 5 x (x - 1) , we get 4 5 x = 4 x - 1 5 x (x - 1) = 4 x - 4 5 x (x - 1) and 3 x - 1 = 3 5 x (x - 1) 5 x = 15 x 5 x (x - 1) .

Now let's do it

4 5 x - 3 x - 1 = 4 x - 4 5 x (x - 1) - 15 x 5 x (x - 1) = 4 x - 4 - 15 x 5 x · (x - 1) = = - 4 - 11 · x 5 · x · (x - 1) = - 4 - 11 · x 5 · x 2 - 5 · x

Answer: - 4 - 11 x 5 x 2 - 5 x

For detailed information, see the article on adding and subtracting algebraic fractions.

Multiplying algebraic fractions

With fractions, you can perform multiplication similar to multiplying ordinary fractions: in order to multiply fractions, you need to multiply the numerators and denominators separately.

Let's look at an example of such a plan.

Example 4

When multiplying 2 x + 2 by x - x · y y from the rule we obtain that 2 x + 2 · x - x · y y = 2 · (x - x · y) (x + 2) · y.

Now you need to perform transformations, that is, multiply a monomial by a polynomial. We get that

2 x - x y (x + 2) y = 2 x - 2 x y x y + 2 y

You must first decompose the fraction into polynomials in order to simplify the fraction. Afterwards you can make a reduction. We have that

2 x 3 - 8 x 3 x y - y 6 y 5 x 2 + 2 x = 2 x (x - 2) (x + 2) y (3 x - 1 ) · 6 · y 5 x · (x + 2) = = 2 · x · (x - 2) · (x + 2) · 6 · y 5 y · (3 · x - 1) · x · x + 2 = 12 · (x - 2) · y 4 3 · x - 1 = 12 · x · y 4 - 24 · y 4 3 · x - 1

A detailed discussion of this action can be found in the article on multiplying and dividing fractions.

Division

Let's look at division with algebraic fractions. Let's apply the rule: in order to divide fractions, you need to multiply the first by the reciprocal of the second.

A fraction that is the reciprocal of a given fraction is considered a fraction with the numerator and denominator reversed. That is, this fraction is called its reciprocal.

Let's look at an example.

Example 5

Divide x 2 - x · y 9 · y 2: 2 · x 3 · y.

Solution

Then the inverse fraction 2 x 3 y will be written as 3 y 2 x. This means that we get that x 2 - x y 9 y 2: 2 x 3 y = x 2 - x y 9 y 2 3 y 2 x = x x - y 3 y 9 · y 2 · 2 · x = x - y 6 · y.

Answer: x 2 - x y 9 y 2: 2 x 3 y = x - y 6 y

Raising an algebraic fraction to a power

If there is a natural power, then it is necessary to apply the rule of actions with raising to a natural power. In such calculations, we use the rule: when raising to a power, you need to separately raise the numerator and denominator to a power, and then write down the result.

Example 6

Let's look at the example of the fraction 2 x x - y. If it is necessary to raise it to a power equal to 2, then we perform the following steps: 2 x x - y 2 = 2 x 2 (x - y) 2. Then we raise the resulting monomial to a power. Having completed the steps, we find that the fraction takes the form 4 x 2 x 2 - 2 x x y + y 2.

A detailed solution to such examples is discussed in the article on raising an algebraic fraction to a power.

When working with fraction powers, remember that the numerator and denominator are raised to powers separately. This significantly simplifies the process of solving and further simplifying the fraction. It is worth paying attention to the sign in front of the degree. If there is a minus sign, then such a fraction should be reversed for ease of calculation.

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At first glance, algebraic fractions seem very complex, and an unprepared student may think that nothing can be done with them. The accumulation of variables, numbers, and even degrees evokes fear. However, the same rules are used to reduce common fractions (such as 15/25) and algebraic fractions.

Steps

Reducing Fractions

Become familiar with operations with simple fractions. Operations with ordinary and algebraic fractions are similar. For example, let's take the fraction 15/35. To simplify this fraction, you should find common divisor. Both numbers are divisible by five, so we can isolate 5 in the numerator and denominator:

15 → 5 * 3 35 → 5 * 7

Now you can reduce common factors, that is, cross out 5 in the numerator and denominator. As a result, we get the simplified fraction 3/7 . In algebraic expressions, common factors are identified in the same way as in ordinary ones. In the previous example we were able to easily isolate 5 from 15 - the same principle applies to more complex expressions such as 15x – 5. Let's find the common factor. In this case it will be 5, since both terms (15x and -5) are divisible by 5. As before, select the common factor and move it left.

15x – 5 = 5 * (3x – 1)

To check if everything is correct, just multiply the expression in brackets by 5 - the result will be the same numbers as at first. Complex members can be isolated in the same way as simple ones. The same principles apply to algebraic fractions as to ordinary ones. This is the easiest way to reduce a fraction. Consider the following fraction:

(x+2)(x-3)(x+2)(x+10)

Note that both the numerator (top) and denominator (bottom) contain a term (x+2), so it can be reduced in the same way as the common factor 5 in the fraction 15/35:

(x+2) (x-3) → (x-3)(x+2) (x+10) → (x+10)

As a result, we get a simplified expression: (x-3)/(x+10)

Reducing algebraic fractions

Find the common factor in the numerator, that is, at the top of the fraction. When reducing an algebraic fraction, the first step is to simplify both sides. Start with the numerator and try to factor it into as many factors as possible. Consider in this section the following fraction:

9x-3 15x+6

Let's start with the numerator: 9x – 3. For 9x and -3, the common factor is the number 3. Let's take 3 out of brackets, as is done with ordinary numbers: 3 * (3x-1). The result of this transformation is the following fraction:

3(3x-1) 15x+6

Find the common factor in the numerator. Let's continue with the example above and write down the denominator: 15x+6. As before, let's find what number both parts are divisible by. And in this case the common factor is 3, so we can write: 3 * (5x +2). Let's rewrite the fraction in the following form:

3(3x-1) 3(5x+2)

Shorten the same terms. At this step you can simplify the fraction. Cancel the same terms in the numerator and denominator. In our example, this number is 3.

3 (3x-1) → (3x-1) 3 (5x+2) → (5x+2)

Determine that the fraction has the simplest form. A fraction is completely simplified when there are no common factors left in the numerator and denominator. Note that you cannot cancel terms that appear inside parentheses - in the example above there is no way to isolate x from 3x and 5x, since the full terms are (3x -1) and (5x + 2). Thus, the fraction cannot be simplified further, and the final answer is as follows:

(3x-1)(5x+2)

Practice reducing fractions on your own. The best way to learn the method is to solve problems yourself. The correct answers are given below the examples.

4(x+2)(x-13)(4x+8)

Answer:(x=13)

2x 2 -x 5x

Answer:(2x-1)/5

Special Moves

Place the negative sign outside the fraction. Suppose you are given the following fraction:

3(x-4) 5(4-x)

Note that (x-4) and (4-x) are “almost” identical, but they cannot be reduced immediately because they are “inverted”. However, (x - 4) can be written as -1 * (4 - x), just as (4 + 2x) can be written as 2 * (2 + x). This is called “sign reversal.”

-1 * 3(4-x) 5(4-x)

Now you can reduce identical terms (4-x):

-1 * 3 (4-x) 5 (4-x)

So, we get the final answer: -3/5 . Learn to recognize the difference between squares. A difference of squares is when the square of one number is subtracted from the square of another number, as in the expression (a 2 - b 2). The difference of perfect squares can always be decomposed into two parts - the sum and the difference of the corresponding square roots. Then the expression will take the following form:

A 2 - b 2 = (a+b)(a-b)

This technique is very useful when finding common terms in algebraic fractions.

Check whether you have correctly factored this or that expression. To do this, multiply the factors - the result should be the same expression.
To completely simplify a fraction, always isolate the largest factors.