C 8 30 fractional rational equations. Fractional-rational equations

Your privacy is important to us. For this reason, we have developed a Privacy Policy that describes how we use and store your information. Please read our privacy policy and let us know if you have any questions.

Collection and use of personal information

Personal information refers to data that can be used to identify a specific person or contact him.

You may be asked to provide your personal information at any time when you contact us.

The following are some examples of the types of personal information we may collect and how we may use such information.

What personal information we collect:

• When you submit an application on the site, we may collect various information, including your name, phone number, email address, etc.

How we use your personal information:

• The personal information we collect allows us to contact you and inform you about unique offers, promotions and other events and upcoming events.
• From time to time, we may use your personal information to send you important notices and messages.
• We may also use personal information for internal purposes, such as conducting audits, data analysis and various research in order to improve the services we provide and provide you with recommendations regarding our services.
• If you enter a prize draw, contest or similar incentive, we may use the information you provide to administer such programs.

Disclosure to third parties

We do not disclose information received from you to third parties.

Exceptions:

• In the event that it is necessary - in accordance with the law, judicial order, in legal proceedings, and / or based on public requests or requests from state bodies in the territory of the Russian Federation - disclose your personal information. We may also disclose information about you if we determine that such disclosure is necessary or appropriate for security, law enforcement, or other public interest purposes.
• In the event of a reorganization, merger or sale, we may transfer the personal information we collect to the relevant third party successor.

Protection of personal information

We take precautions - including administrative, technical and physical - to protect your personal information from loss, theft, and misuse, as well as from unauthorized access, disclosure, alteration and destruction.

Maintaining your privacy at the company level

To ensure that your personal information is secure, we communicate privacy and security practices to our employees and strictly enforce privacy practices.

We introduced the equation above in § 7. First, we recall what a rational expression is. This is an algebraic expression made up of numbers and the variable x using the operations of addition, subtraction, multiplication, division and exponentiation with a natural exponent.

If r(x) is a rational expression, then the equation r(x) = 0 is called a rational equation.

However, in practice it is more convenient to use a somewhat broader interpretation of the term "rational equation": this is an equation of the form h(x) = q(x), where h(x) and q(x) are rational expressions.

Until now, we could not solve any rational equation, but only one that, as a result of various transformations and reasoning, was reduced to linear equation. Now our possibilities are much greater: we will be able to solve a rational equation, which reduces not only to linear
mu, but also to the quadratic equation.

Recall how we solved rational equations earlier and try to formulate a solution algorithm.

Example 1 solve the equation

Decision. We rewrite the equation in the form

In this case, as usual, we use the fact that the equalities A \u003d B and A - B \u003d 0 express the same relationship between A and B. This allowed us to transfer the term to the left side of the equation with the opposite sign.

Let's perform transformations of the left side of the equation. We have

Recall the equality conditions fractions zero: if, and only if, two relations are simultaneously satisfied:

1) the numerator of the fraction is zero (a = 0); 2) the denominator of the fraction is different from zero).
Equating to zero the numerator of the fraction on the left side of equation (1), we obtain

It remains to check the fulfillment of the second condition mentioned above. The ratio means for equation (1) that . The values ​​x 1 = 2 and x 2 = 0.6 satisfy the indicated relationships and therefore serve as the roots of equation (1), and at the same time the roots of the given equation.

1) Let's transform the equation to the form

2) Let's perform the transformations of the left side of this equation:

(simultaneously changed the signs in the numerator and
fractions).
Thus, the given equation takes the form

3) Solve the equation x 2 - 6x + 8 = 0. Find

4) For the found values, check the condition . The number 4 satisfies this condition, but the number 2 does not. So 4 is the root of the given equation, and 2 is an extraneous root.
Answer: 4.

2. Solution of rational equations by introducing a new variable

The method of introducing a new variable is familiar to you, we have used it more than once. Let us show by examples how it is used in solving rational equations.

Example 3 Solve the equation x 4 + x 2 - 20 = 0.

Decision. We introduce a new variable y \u003d x 2. Since x 4 \u003d (x 2) 2 \u003d y 2, then the given equation can be rewritten in the form

y 2 + y - 20 = 0.

This is a quadratic equation, the roots of which we will find using the known formulas; we get y 1 = 4, y 2 = - 5.
But y \u003d x 2, which means that the problem has been reduced to solving two equations:
x2=4; x 2 \u003d -5.

From the first equation we find the second equation has no roots.
Answer: .
An equation of the form ax 4 + bx 2 + c \u003d 0 is called a biquadratic equation (“bi” - two, i.e., as it were, a “twice square” equation). The equation just solved was exactly biquadratic. Any biquadratic equation is solved in the same way as the equation from example 3: a new variable y \u003d x 2 is introduced, the resulting quadratic equation is solved with respect to the variable y, and then returned to the variable x.

Example 4 solve the equation

Decision. Note that the same expression x 2 + 3x occurs twice here. Hence, it makes sense to introduce a new variable y = x 2 + Zx. This will allow us to rewrite the equation in a simpler and more pleasant form (which, in fact, is the purpose of introducing a new variable- and recording is easier
, and the structure of the equation becomes clearer):

And now we will use the algorithm for solving a rational equation.

1) Let's move all the terms of the equation into one part:

= 0
2) Let's transform the left side of the equation

So, we have transformed the given equation into the form

3) From the equation - 7y 2 + 29y -4 = 0 we find (we have already solved quite a lot of quadratic equations, so it’s probably not worth always giving detailed calculations in the textbook).

4) Let's check the found roots using the condition 5 (y - 3) (y + 1). Both roots satisfy this condition.
So, the quadratic equation for the new variable y is solved:
Since y \u003d x 2 + Zx, and y, as we have established, takes two values: 4 and, - we still have to solve two equations: x 2 + Zx \u003d 4; x 2 + Zx \u003d. The roots of the first equation are the numbers 1 and - 4, the roots of the second equation are the numbers

In the examples considered, the method of introducing a new variable was, as mathematicians like to say, adequate to the situation, that is, it corresponded well to it. Why? Yes, because the same expression was clearly encountered in the equation several times and it was reasonable to designate this expression with a new letter. But this is not always the case, sometimes a new variable "appears" only in the process of transformations. This is exactly what will happen in the next example.

Example 5 solve the equation
x(x-1)(x-2)(x-3) = 24.
Decision. We have
x (x - 3) \u003d x 2 - 3x;
(x - 1) (x - 2) \u003d x 2 -3x + 2.

So the given equation can be rewritten as

(x 2 - 3x)(x 2 + 3x + 2) = 24

Now a new variable has "appeared": y = x 2 - Zx.

With its help, the equation can be rewritten in the form y (y + 2) \u003d 24 and then y 2 + 2y - 24 \u003d 0. The roots of this equation are the numbers 4 and -6.

Returning to the original variable x, we obtain two equations x 2 - Zx \u003d 4 and x 2 - Zx \u003d - 6. From the first equation we find x 1 \u003d 4, x 2 \u003d - 1; the second equation has no roots.

Answer: 4, - 1.

Lesson content lesson summary support frame lesson presentation accelerative methods interactive technologies Practice tasks and exercises self-examination workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures graphics, tables, schemes humor, anecdotes, jokes, comics parables, sayings, crossword puzzles, quotes Add-ons abstracts articles chips for inquisitive cheat sheets textbooks basic and additional glossary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in the textbook elements of innovation in the lesson replacing obsolete knowledge with new ones Only for teachers perfect lessons calendar plan for the year methodological recommendations of the discussion program Integrated Lessons

Simply put, these are equations in which there is at least one with a variable in the denominator.

For example:

\(\frac(9x^2-1)(3x)\) \(=0\)
\(\frac(1)(2x)+\frac(x)(x+1)=\frac(1)(2)\)
\(\frac(6)(x+1)=\frac(x^2-5x)(x+1)\)

Example not fractional rational equations:

\(\frac(9x^2-1)(3)\) \(=0\)
\(\frac(x)(2)\) \(+8x^2=6\)

How are fractional rational equations solved?

The main thing to remember about fractional rational equations is that you need to write in them. And after finding the roots, be sure to check them for admissibility. Otherwise, extraneous roots may appear, and the whole solution will be considered incorrect.

Algorithm for solving a fractional rational equation:

Write out and "solve" the ODZ.

Multiply each term in the equation by a common denominator and reduce the resulting fractions. The denominators will disappear.

Write the equation without opening brackets.

Solve the resulting equation.

Check the found roots with ODZ.

Write down in response the roots that passed the test in step 7.

Do not memorize the algorithm, 3-5 solved equations - and it will be remembered by itself.

Example . Solve fractional rational equation \(\frac(x)(x-2) - \frac(7)(x+2)=\frac(8)(x^2-4)\)

Decision:

Answer: \(3\).

Example . Find the roots of the fractional rational equation \(=0\)

Decision:

\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)(x^2+7x+10)\)\(=0\) ODZ: \(x+2≠0⇔x≠-2\) \(x+5≠0 ⇔x≠-5\) \(x^2+7x+10≠0\) \(D=49-4 \cdot 10=9\) \(x_1≠\frac(-7+3)(2)=-2\) \(x_2≠\frac(-7-3)(2)=-5\)		We write down and "solve" ODZ. Expand \(x^2+7x+10\) into the formula: \(ax^2+bx+c=a(x-x_1)(x-x_2)\). Fortunately \(x_1\) and \(x_2\) we have already found.
\(\frac(x)(x+2) + \frac(x+1)(x+5)-\frac(7-x)((x+2)(x+5))\)\(=0\)		Obviously, the common denominator of fractions: \((x+2)(x+5)\). We multiply the whole equation by it.
\(\frac(x(x+2)(x+5))(x+2) + \frac((x+1)(x+2)(x+5))(x+5)-\) \(-\frac((7-x)(x+2)(x+5))((x+2)(x+5))\)\(=0\)		We reduce fractions
\(x(x+5)+(x+1)(x+2)-7+x=0\)		Opening the brackets
\(x^2+5x+x^2+3x+2-7+x=0\)		We give like terms
\(2x^2+9x-5=0\)		Finding the roots of the equation
\(x_1=-5;\) \(x_2=\frac(1)(2).\)		One of the roots does not fit under the ODZ, so in response we write down only the second root.

Answer: \(\frac(1)(2)\).

We continue talking about solution of equations. In this article, we will focus on rational equations and principles for solving rational equations with one variable. First, let's figure out what kind of equations are called rational, give a definition of integer rational and fractional rational equations, and give examples. Next, we will obtain algorithms for solving rational equations, and, of course, consider the solutions of typical examples with all the necessary explanations.

Page navigation.

Based on the sounded definitions, we give several examples of rational equations. For example, x=1 , 2 x−12 x 2 y z 3 =0 , , are all rational equations.

From the examples shown, it can be seen that rational equations, as well as equations of other types, can be either with one variable, or with two, three, etc. variables. In the following paragraphs, we will talk about solving rational equations in one variable. Solving equations with two variables and their large number deserve special attention.

In addition to dividing rational equations by the number of unknown variables, they are also divided into integer and fractional. Let us give the corresponding definitions.

Definition.

The rational equation is called whole, if both its left and right sides are integer rational expressions.

Definition.

If at least one of the parts of a rational equation is a fractional expression, then such an equation is called fractionally rational(or fractional rational).

It is clear that integer equations do not contain division by a variable; on the contrary, fractional rational equations necessarily contain division by a variable (or a variable in the denominator). So 3 x+2=0 and (x+y) (3 x 2 −1)+x=−y+0.5 are entire rational equations, both of their parts are integer expressions. A and x:(5 x 3 +y 2)=3:(x−1):5 are examples of fractional rational equations.

Concluding this paragraph, let us pay attention to the fact that the linear equations and quadratic equations known by this moment are entire rational equations.

Solving integer equations

One of the main approaches to solving entire equations is their reduction to equivalent ones algebraic equations. This can always be done by performing the following equivalent transformations of the equation:

• first, the expression from the right side of the original integer equation is transferred to the left side with the opposite sign to get zero on the right side;
• after that, on the left side of the equation, the resulting standard form.

The result is an algebraic equation that is equivalent to the original whole equation. So in the simplest cases, the solution of entire equations is reduced to the solution of linear or quadratic equations, and in the general case - to the solution of an algebraic equation of degree n. For clarity, let's analyze the solution of the example.

Example.

Find the roots of the whole equation 3 (x+1) (x−3)=x (2 x−1)−3.

Decision.

Let us reduce the solution of this whole equation to the solution of an equivalent algebraic equation. To do this, firstly, we transfer the expression from the right side to the left, as a result we arrive at the equation 3 (x+1) (x−3)−x (2 x−1)+3=0. And, secondly, we transform the expression formed on the left side into a polynomial of the standard form by doing the necessary: 3 (x+1) (x−3)−x (2 x−1)+3= (3 x+3) (x−3)−2 x 2 +x+3= 3 x 2 −9 x+3 x−9−2 x 2 +x+3=x 2 −5 x−6. Thus, the solution of the original integer equation is reduced to the solution of the quadratic equation x 2 −5·x−6=0 .

Calculate its discriminant D=(−5) 2 −4 1 (−6)=25+24=49, it is positive, which means that the equation has two real roots, which we find by the formula of the roots of the quadratic equation:

To be completely sure, let's do checking the found roots of the equation. First, we check the root 6, substitute it instead of the variable x in the original integer equation: 3 (6+1) (6−3)=6 (2 6−1)−3, which is the same, 63=63 . This is a valid numerical equation, so x=6 is indeed the root of the equation. Now we check the root −1 , we have 3 (−1+1) (−1−3)=(−1) (2 (−1)−1)−3, whence, 0=0 . For x=−1, the original equation also turned into a true numerical equality, therefore, x=−1 is also the root of the equation.

Answer:

6 , −1 .

Here it should also be noted that the term “power of an entire equation” is associated with the representation of an entire equation in the form of an algebraic equation. We give the corresponding definition:

Definition.

The degree of the whole equation call the degree of an algebraic equation equivalent to it.

According to this definition, the whole equation from the previous example has the second degree.

On this one could finish with the solution of entire rational equations, if not for one but .... As is known, the solution of algebraic equations of degree higher than the second is associated with significant difficulties, and for equations of degree higher than the fourth, there are no general formulas for roots at all. Therefore, to solve entire equations of the third, fourth, and higher degrees, one often has to resort to other solution methods.

In such cases, sometimes the approach to solving entire rational equations based on factorization method. At the same time, the following algorithm is followed:

• first, they strive to have zero on the right side of the equation, for this they transfer the expression from the right side of the whole equation to the left;
• then, the resulting expression on the left side is presented as a product of several factors, which allows you to go to a set of several simpler equations.

The above algorithm for solving the whole equation through factorization requires a detailed explanation using an example.

Example.

Solve the whole equation (x 2 −1) (x 2 −10 x+13)= 2 x (x 2 −10 x+13) .

Decision.

First, as usual, we transfer the expression from the right side to the left side of the equation, not forgetting to change the sign, we get (x 2 −1) (x 2 −10 x+13) − 2 x (x 2 −10 x+13)=0 . It is quite obvious here that it is not advisable to transform the left side of the resulting equation into a polynomial of the standard form, since this will give an algebraic equation of the fourth degree of the form x 4 −12 x 3 +32 x 2 −16 x−13=0, whose solution is difficult.

On the other hand, it is obvious that x 2 −10·x+13 can be found on the left side of the resulting equation, thereby representing it as a product. We have (x 2 −10 x+13) (x 2 −2 x−1)=0. The resulting equation is equivalent to the original whole equation, and it, in turn, can be replaced by a set of two quadratic equations x 2 −10·x+13=0 and x 2 −2·x−1=0 . Finding their roots using the known root formulas through the discriminant is not difficult, the roots are equal. They are the desired roots of the original equation.

Answer:

It is also useful for solving entire rational equations. method for introducing a new variable. In some cases, it allows one to pass to equations whose degree is lower than the degree of the original integer equation.

Example.

Find the real roots of a rational equation (x 2 +3 x+1) 2 +10=−2 (x 2 +3 x−4).

Decision.

Reducing this whole rational equation to an algebraic equation is, to put it mildly, not a very good idea, since in this case we will come to the need to solve a fourth-degree equation that does not have rational roots. Therefore, you will have to look for another solution.

It is easy to see here that you can introduce a new variable y and replace the expression x 2 +3 x with it. Such a replacement leads us to the whole equation (y+1) 2 +10=−2 (y−4) , which, after transferring the expression −2 (y−4) to the left side and subsequent transformation of the expression formed there, reduces to equation y 2 +4 y+3=0 . The roots of this equation y=−1 and y=−3 are easy to find, for example, they can be found based on the inverse theorem of Vieta's theorem.

Now let's move on to the second part of the method of introducing a new variable, that is, to making a reverse substitution. After performing the reverse substitution, we obtain two equations x 2 +3 x=−1 and x 2 +3 x=−3 , which can be rewritten as x 2 +3 x+1=0 and x 2 +3 x+3 =0 . According to the formula of the roots of the quadratic equation, we find the roots of the first equation. And the second quadratic equation has no real roots, since its discriminant is negative (D=3 2 −4 3=9−12=−3 ).

Answer:

In general, when we are dealing with integer equations of high degrees, we must always be ready to look for a non-standard method or an artificial technique for solving them.

Solution of fractionally rational equations

First, it will be useful to understand how to solve fractionally rational equations of the form , where p(x) and q(x) are rational integer expressions. And then we will show how to reduce the solution of the remaining fractionally rational equations to the solution of equations of the indicated form.

One of the approaches to solving the equation is based on the following statement: the numerical fraction u / v, where v is a non-zero number (otherwise we will encounter , which is not defined), is zero if and only if its numerator is zero, then is, if and only if u=0 . By virtue of this statement, the solution of the equation is reduced to the fulfillment of two conditions p(x)=0 and q(x)≠0 .

This conclusion is consistent with the following algorithm for solving a fractionally rational equation. To solve a fractional rational equation of the form

• solve the whole rational equation p(x)=0 ;
• and check whether the condition q(x)≠0 is satisfied for each found root, while
• if true, then this root is the root of the original equation;
• if not, then this root is extraneous, that is, it is not the root of the original equation.

Let's analyze an example of using the voiced algorithm when solving a fractional rational equation.

Example.

Find the roots of the equation.

Decision.

This is a fractionally rational equation of the form , where p(x)=3 x−2 , q(x)=5 x 2 −2=0 .

According to the algorithm for solving fractionally rational equations of this kind, we first need to solve the equation 3·x−2=0 . This is a linear equation whose root is x=2/3 .

It remains to check for this root, that is, to check whether it satisfies the condition 5·x 2 −2≠0 . We substitute the number 2/3 instead of x into the expression 5 x 2 −2, we get . The condition is met, so x=2/3 is the root of the original equation.

Answer:

2/3 .

The solution of a fractional rational equation can be approached from a slightly different position. This equation is equivalent to the whole equation p(x)=0 on the variable x of the original equation. That is, you can follow this algorithm for solving a fractionally rational equation :

• solve the equation p(x)=0 ;
• find ODZ variable x ;
• take the roots belonging to the region of admissible values ​​- they are the desired roots of the original fractional rational equation.

For example, let's solve a fractional rational equation using this algorithm.

Example.

Solve the equation.

Decision.

First, we solve the quadratic equation x 2 −2·x−11=0 . Its roots can be calculated using the root formula for an even second coefficient, we have D 1 =(−1) 2 −1 (−11)=12, and .

Secondly, we find the ODZ of the variable x for the original equation. It consists of all numbers for which x 2 +3 x≠0 , which is the same x (x+3)≠0 , whence x≠0 , x≠−3 .

It remains to check whether the roots found at the first step are included in the ODZ. Obviously yes. Therefore, the original fractionally rational equation has two roots.

Answer:

Note that this approach is more profitable than the first one if the ODZ is easily found, and it is especially beneficial if the roots of the equation p(x)=0 are irrational, for example, , or rational, but with a rather large numerator and/or denominator, for example, 127/1101 and -31/59 . This is due to the fact that in such cases, checking the condition q(x)≠0 will require significant computational efforts, and it is easier to exclude extraneous roots from the ODZ.

In other cases, when solving the equation, especially when the roots of the equation p(x)=0 are integers, it is more advantageous to use the first of the above algorithms. That is, it is advisable to immediately find the roots of the whole equation p(x)=0 , and then check whether the condition q(x)≠0 is satisfied for them, and not find the ODZ, and then solve the equation p(x)=0 on this ODZ . This is due to the fact that in such cases it is usually easier to make a check than to find the ODZ.

Consider the solution of two examples to illustrate the stipulated nuances.

Example.

Find the roots of the equation.

Decision.

First we find the roots of the whole equation (2 x−1) (x−6) (x 2 −5 x+14) (x+1)=0, compiled using the numerator of the fraction. The left side of this equation is a product, and the right side is zero, therefore, according to the method of solving equations through factorization, this equation is equivalent to the set of four equations 2 x−1=0 , x−6=0 , x 2 −5 x+ 14=0 , x+1=0 . Three of these equations are linear and one is quadratic, we can solve them. From the first equation we find x=1/2, from the second - x=6, from the third - x=7, x=−2, from the fourth - x=−1.

With the roots found, it is quite easy to check them to see if the denominator of the fraction on the left side of the original equation does not vanish, and it is not so easy to determine the ODZ, since this will have to solve an algebraic equation of the fifth degree. Therefore, we will refuse to find the ODZ in favor of checking the roots. To do this, we substitute them in turn instead of the variable x in the expression x 5 −15 x 4 +57 x 3 −13 x 2 +26 x+112, obtained after substitution, and compare them with zero: (1/2) 5 −15 (1/2) 4 + 57 (1/2) 3 −13 (1/2) 2 +26 (1/2)+112= 1/32−15/16+57/8−13/4+13+112= 122+1/32≠0 ;
6 5 −15 6 4 +57 6 3 −13 6 2 +26 6+112= 448≠0 ;
7 5 −15 7 4 +57 7 3 −13 7 2 +26 7+112=0;
(−2) 5 −15 (−2) 4 +57 (−2) 3 −13 (−2) 2 + 26 (−2)+112=−720≠0 ;
(−1) 5 −15 (−1) 4 +57 (−1) 3 −13 (−1) 2 + 26·(−1)+112=0 .

Thus, 1/2, 6 and −2 are the desired roots of the original fractionally rational equation, and 7 and −1 are extraneous roots.

Answer:

1/2 , 6 , −2 .

Example.

Find the roots of a fractional rational equation.

Decision.

First we find the roots of the equation (5x2 −7x−1)(x−2)=0. This equation is equivalent to a set of two equations: the square 5·x 2 −7·x−1=0 and the linear x−2=0 . According to the formula of the roots of the quadratic equation, we find two roots, and from the second equation we have x=2.

Checking if the denominator does not vanish at the found values ​​of x is rather unpleasant. And to determine the range of acceptable values ​​of the variable x in the original equation is quite simple. Therefore, we will act through the ODZ.

In our case, the ODZ of the variable x of the original fractional rational equation is made up of all numbers, except for those for which the condition x 2 +5·x−14=0 is satisfied. The roots of this quadratic equation are x=−7 and x=2, from which we conclude about the ODZ: it is made up of all x such that .

It remains to check whether the found roots and x=2 belong to the region of admissible values. The roots - belong, therefore, they are the roots of the original equation, and x=2 does not belong, therefore, it is an extraneous root.

Answer:

It will also be useful to dwell separately on cases where a fractional rational equation of the form contains a number in the numerator, that is, when p (x) is represented by some number. Wherein

• if this number is different from zero, then the equation has no roots, since the fraction is zero if and only if its numerator is zero;
• if this number is zero, then the root of the equation is any number from the ODZ.

Example.

Decision.

Since there is a non-zero number in the numerator of the fraction on the left side of the equation, for no x can the value of this fraction be equal to zero. Therefore, this equation has no roots.

Answer:

no roots.

Example.

Solve the equation.

Decision.

The numerator of the fraction on the left side of this fractional rational equation is zero, so the value of this fraction is zero for any x for which it makes sense. In other words, the solution to this equation is any value of x from the DPV of this variable.

It remains to determine this range of acceptable values. It includes all such values ​​x for which x 4 +5 x 3 ≠0. The solutions of the equation x 4 +5 x 3 \u003d 0 are 0 and −5, since this equation is equivalent to the equation x 3 (x + 5) \u003d 0, and it, in turn, is equivalent to the combination of two equations x 3 \u003d 0 and x +5=0 , from where these roots are visible. Therefore, the desired range of acceptable values ​​are any x , except for x=0 and x=−5 .

Thus, a fractionally rational equation has infinitely many solutions, which are any numbers except zero and minus five.

Answer:

Finally, it's time to talk about solving arbitrary fractional rational equations. They can be written as r(x)=s(x) , where r(x) and s(x) are rational expressions, and at least one of them is fractional. Looking ahead, we say that their solution is reduced to solving equations of the form already familiar to us.

It is known that the transfer of a term from one part of the equation to another with the opposite sign leads to an equivalent equation, so the equation r(x)=s(x) is equivalent to the equation r(x)−s(x)=0 .

We also know that any can be identically equal to this expression. Thus, we can always transform the rational expression on the left side of the equation r(x)−s(x)=0 into an identically equal rational fraction of the form .

So we go from the original fractional rational equation r(x)=s(x) to the equation , and its solution, as we found out above, reduces to solving the equation p(x)=0 .

But here it is necessary to take into account the fact that when replacing r(x)−s(x)=0 with , and then with p(x)=0 , the range of allowable values ​​of the variable x may expand.

Therefore, the original equation r(x)=s(x) and the equation p(x)=0 , which we arrived at, may not be equivalent, and by solving the equation p(x)=0 , we can get roots that will be extraneous roots of the original equation r(x)=s(x) . It is possible to identify and not include extraneous roots in the answer, either by checking, or by checking their belonging to the ODZ of the original equation.

We summarize this information in algorithm for solving a fractional rational equation r(x)=s(x). To solve the fractional rational equation r(x)=s(x) , one must

• Get zero on the right by moving the expression from the right side with the opposite sign.
• Perform actions with fractions and polynomials on the left side of the equation, thereby converting it into a rational fraction of the form.
• Solve the equation p(x)=0 .
• Identify and exclude extraneous roots, which is done by substituting them into the original equation or by checking their belonging to the ODZ of the original equation.

For greater clarity, we will show the entire chain of solving fractional rational equations:
.

Let's go through the solutions of several examples with a detailed explanation of the solution in order to clarify the given block of information.

Example.

Solve a fractional rational equation.

Decision.

We will act in accordance with the just obtained solution algorithm. And first we transfer the terms from the right side of the equation to the left side, as a result we pass to the equation .

In the second step, we need to convert the fractional rational expression on the left side of the resulting equation to the form of a fraction. To do this, we perform the reduction of rational fractions to a common denominator and simplify the resulting expression: . So we come to the equation.

In the next step, we need to solve the equation −2·x−1=0 . Find x=−1/2 .

It remains to check whether the found number −1/2 is an extraneous root of the original equation. To do this, you can check or find the ODZ variable x of the original equation. Let's demonstrate both approaches.

Let's start with a check. We substitute the number −1/2 instead of the variable x into the original equation, we get , which is the same, −1=−1. The substitution gives the correct numerical equality, therefore, x=−1/2 is the root of the original equation.

Now we will show how the last step of the algorithm is performed through the ODZ. The range of admissible values ​​of the original equation is the set of all numbers except −1 and 0 (when x=−1 and x=0, the denominators of fractions vanish). The root x=−1/2 found at the previous step belongs to the ODZ, therefore, x=−1/2 is the root of the original equation.

Answer:

−1/2 .

Let's consider another example.

Example.

Find the roots of the equation.

Decision.

We need to solve a fractionally rational equation, let's go through all the steps of the algorithm.

First, we transfer the term from the right side to the left, we get .

Secondly, we transform the expression formed on the left side: . As a result, we arrive at the equation x=0 .

Its root is obvious - it is zero.

At the fourth step, it remains to find out if the root found is not an outside one for the original fractionally rational equation. When it is substituted into the original equation, the expression is obtained. Obviously, it does not make sense, since it contains division by zero. Whence we conclude that 0 is an extraneous root. Therefore, the original equation has no roots.

7 , which leads to the equation . From this we can conclude that the expression in the denominator of the left side must be equal to the right side, that is, . Now we subtract from both parts of the triple: . By analogy, from where, and further.

The check shows that both found roots are the roots of the original fractional rational equation.

Answer:

Bibliography.

• Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.

An integer expression is a mathematical expression made up of numbers and literal variables using the operations of addition, subtraction, and multiplication. Integers also include expressions that include division by some number other than zero.

The concept of a fractional rational expression

A fractional expression is a mathematical expression that, in addition to the operations of addition, subtraction and multiplication performed with numbers and literal variables, as well as division by a number not equal to zero, also contains division into expressions with literal variables.

Rational expressions are all integer and fractional expressions. Rational equations are equations whose left and right sides are rational expressions. If in a rational equation the left and right parts are integer expressions, then such a rational equation is called an integer.

If in a rational equation the left or right parts are fractional expressions, then such a rational equation is called fractional.

Examples of fractional rational expressions

1.x-3/x=-6*x+19

2. (x-4)/(2*x+5) = (x+7)/(x-2)

3. (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5))

Scheme for solving a fractional rational equation

1. Find the common denominator of all fractions that are included in the equation.

2. Multiply both sides of the equation by a common denominator.

3. Solve the resulting whole equation.

4. Check the roots, and exclude those that turn the common denominator to zero.

Since we are solving fractional rational equations, there will be variables in the denominators of the fractions. So, they will be in a common denominator. And in the second paragraph of the algorithm, we multiply by a common denominator, then extraneous roots may appear. At which the common denominator will be equal to zero, which means that multiplication by it will be meaningless. Therefore, at the end, be sure to check the obtained roots.

Consider an example:

Solve a fractional rational equation: (x-3)/(x-5) + 1/x = (x+5)/(x*(x-5)).

We will adhere to the general scheme: we first find the common denominator of all fractions. We get x*(x-5).

Multiply each fraction by a common denominator and write the resulting whole equation.

(x-3)/(x-5) * (x*(x-5))= x*(x+3);
1/x * (x*(x-5)) = (x-5);
(x+5)/(x*(x-5)) * (x*(x-5)) = (x+5);
x*(x+3) + (x-5) = (x+5);

Let's simplify the resulting equation. We get:

x^2+3*x + x-5 - x - 5 =0;
x^2+3*x-10=0;

We got a simple reduced quadratic equation. We solve it by any of the known methods, we get the roots x=-2 and x=5.

Now we check the obtained solutions:

We substitute the numbers -2 and 5 in the common denominator. At x=-2, the common denominator x*(x-5) does not vanish, -2*(-2-5)=14. So the number -2 will be the root of the original fractional rational equation.

At x=5, the common denominator x*(x-5) becomes zero. Therefore, this number is not the root of the original fractional rational equation, since there will be division by zero.