What is a rational fraction examples. Rational fraction

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

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

Actually, according to the formula for abbreviated multiplication: they allow you to collapse a polynomial into one or more factors;
Using the factorization of a quadratic trinomial through a discriminant. The same method makes it possible to verify that any trinomial cannot be factorized at all;
The grouping method is the most complex tool, but it is the only way, which works if the previous two did not work.

As you may have guessed from the title of this video, we will again talk about rational fractions. Just a few minutes ago, I finished a lesson with one tenth grader, and there we analyzed exactly these expressions. Therefore, this lesson will be intended specifically for high school students.

Surely many now have a question: “Why should students in grades 10-11 study such simple things as rational fractions, because this is taught in grade 8?” But the trouble is that most people “go through” this topic. In the 10th-11th grade, they no longer remember how to do multiplication, division, subtraction and addition of rational fractions from the 8th grade, and yet these simple knowledge further, more are being built complex designs, as a solution to logarithmic, trigonometric equations and many other 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;
$((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2))$ — square of the sum or difference;
$((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b)^( 2)) \right)$ is the sum of cubes;
$((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^(2) ) \right)$ is the difference of cubes.

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

The second, completely obvious formula is the decomposition quadratic trinomial by multipliers:

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

WITH theoretical part we figured it out. But how to solve real rational fractions, which are covered in 8th grade? Now we will practice.

Task No. 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 factorization is needed at all. The fact is that at first glance at the first part of the task, you want to reduce the cube with the square, but this is strictly forbidden, because they are terms in the numerator and denominator, but in no case are factors.

What is abbreviation anyway? Reduction is the use of a basic rule for working 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 in this case, when we reduce, we, on the contrary, 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 reduce the numerator with the denominator only when both of them are factorized. Let's do this.

Now you need to see how many terms are in a particular element, and accordingly find out which formula to use.

Let's transform each expression into 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 look at the denominator. Let's expand it using the difference of squares formula:

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

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 entire structure 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 conclusion from these constructions is the following:

Not every polynomial can be factorized.
Even if it is decomposed, you need to carefully look at what exactly the abbreviated multiplication formula is.

To do this, firstly, we need to estimate how many terms there are (if there are two, 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, then this , uniquely, 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 No. 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: look at its numerator and make possible transformations:

Now let's look at the denominator:

With the second fraction: nothing can be done at all in the numerator, because it linear expression, and it is impossible to remove any factor from it. Let's look at the denominator:

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

Let's go to the third fraction. Numerator:

Let's look at the denominator of the last fraction:

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

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

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

Nuances of the solution

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

Problem No. 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 look at the first part:

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

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

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

Let's rewrite the original expression:

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

Now let's look at 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 two elements could not be grouped, we grouped three. All that remains is 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:

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

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

Nuances of the solution

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

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

Problem No. 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 figure it out 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:

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

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

We will write the numerator separately:

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

Consequently, this polynomial cannot be factorized.

We have already done the maximum that we could do and decompose.

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, problem solved.

To be honest, it wasn't that great difficult task: everything was easily factored there and quickly reduced similar terms, and everything was shrinking beautifully. So now let's try to solve a more serious problem.

Problem No. 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 bracket. From the very beginning, let’s factorize 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:

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

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

We return to our original design 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

Again key facts today's video lesson:

You need to know by heart the formulas for abbreviated multiplication - and not just know, but be able to see in those expressions that you will encounter in real problems. A wonderful rule can help us with this: if there are two terms, then it 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 any construction cannot be expanded using abbreviated multiplication formulas, then either standard formula factorization of trinomials, or the grouping method.
If something doesn’t work out, look carefully at the source expression to see if any transformations are required with it at all. Perhaps it will be enough to simply put the factor out of brackets, 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 lead to common denominator, and only after that, when all the fractions have been reduced to it, be sure to bring the same in the new numerator, and then factor the new numerator again - perhaps something will be reduced.

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

Definition.The sum of integer non-negative powers of an unknown X, taken with certain numerical coefficients, is called a polynomial.

Here: - real numbers.

n- degree of the polynomial.

Operations on polynomials.

1). When adding (subtracting) two polynomials, the coefficients are added (subtracted) equal degrees unknown x.

2). Two polynomials are equal if they have the same degree and equal coefficients at the same powers of X.

3). The degree of a polynomial obtained by multiplying two polynomials is equal to the sum of the degrees of the polynomials being multiplied.

4). Linear operations on polynomials have the properties of associativity, commutativity and distributivity.

5) The division of a polynomial by a polynomial can be done using the “division by a corner” rule.

Definition. The number x=a is called the root of a polynomial if its substitution into a polynomial turns it into zero, i.e.

Bezout's theorem. Polynomial remainder
by binomial (x-a) is equal to the value of the polynomial at x=a, i.e.

Proof.

Let where

Putting x=a in the equality, we get

1). When dividing a polynomial by a binomial (x-a), the remainder will always be a number.

2). If a is the root of a polynomial, then the polynomial is divisible by the binomial (x-a) without a remainder.

3) When dividing a polynomial of degree n by a binomial (x-a), we obtain a polynomial of degree (n-1).

Fundamental theorem of algebra.Any polynomial of degreen (n>1) has at least one root(presented without proof).

Consequence.Any polynomial of degree n has exactly n roots and over the field of complex numbers is decomposed into the product n linear factors, i.e. Among the roots of the polynomial there may be repeating numbers (multiple roots). For polynomials with real coefficients, complex roots can appear only in conjugate pairs. Let us prove the last statement.

Let
- complex root polynomial, then Based on general property complex numbers can be stated therefore
- also a root.

Each pair of complex conjugate roots of a polynomial corresponds to a square trinomial with real coefficients.

Here p, q- real numbers (show example).

Conclusion.We can represent any polynomial as a product of linear factors and square trinomials with real coefficients.

Rational fractions.

A rational fraction is the ratio of two polynomials.

If
, then the rational fraction is called proper. IN otherwise the fraction is incorrect. Any improper fraction can be represented as the sum of a polynomial (quotient) and a proper rational fraction by dividing the polynomial in the numerator by the polynomial in the denominator.

- improper rational fraction.

This improper rational fraction can now be represented in the following form.

Taking into account what has been shown, in the future we will consider only proper rational fractions.

There are so-called simple rational fractions - these are fractions that cannot be simplified in any way. These simplest fractions look like:

A proper rational fraction of a more complex form can always be represented as a sum of the simplest rational fractions. The set of fractions is determined by the set of roots of the polynomial that appears in the denominator of a proper irreducible rational fraction. The rule for decomposing a fraction into its simplest is as follows.

Let the rational fraction be represented in the following form.

Here, the numerator of the simplest fractions contains unknown coefficients, which can always be determined by the method of undetermined coefficients. The essence of the method is to equate the coefficients at the same powers of X for the polynomial in the numerator of the original fraction and the polynomial in the numerator of the fraction obtained after reducing the simplest fractions to a common denominator.

Let us equate the coefficients for the same powers of X.

Solving the system of equations for unknown coefficients, we obtain.

So, this fraction can be represented by a set of the following simple fractions.

By bringing it to a common denominator, we are convinced of the correctness of the solution to the problem.

She looks like

where P(x) and Q(x) are some polynomials.

Distinguish between proper and improper rational fractions, by analogy with ordinary fractions numerical fractions. A rational fraction is called proper if the order of the denominator is more order numerator, and incorrect if vice versa.

Any improper rational fraction can be converted into the sum of some polynomial and a proper rational fraction

Any rational fraction of polynomials with real coefficients can be represented as a sum of rational fractions whose denominators are the expressions (x − a) k (a is the real root of Q(x)) or (x 2 + px + q) k (Where x 2 + px + q doesn't have real roots), and the degree k is not greater than the multiplicity corresponding roots in the polynomial Q(x). Based on this statement, the theorem on the integrability of rational fractions is based. According to it, any rational fraction can be integrated into elementary functions, which makes the class of rational fractions very important in mathematical analysis.

What is a rational fraction examples. Rational fraction

Formulas for solving problems

Task No. 1

Nuances of multiplying rational fractions

Problem No. 2

Nuances of the solution

Problem No. 3

Nuances of the solution

Problem No. 4

Problem No. 5

Key points

Rational fractions.

see also

See what “Rational fraction” is in other dictionaries: