Fractionally rational equations of definition with the authors. Rational Equations

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. Further, we will obtain algorithms for solving rational equations, and, of course, consider the solutions of typical examples with all the necessary explanations.

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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 them a 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 parts 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 linear equations and quadratic equations known by this moment are entire rational equations.

Solving entire equations

One of the main approaches to solving entire equations is their reduction to equivalent 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 view.

The result is algebraic equation, which 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 more high degrees often have to resort to other methods of solution.

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 seek 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 known formulas roots 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 integer 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 equal to zero if and only if its numerator zero, that 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 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 number is in the numerator in a fractional rational equation of the form, 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. Hence, given 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 came to, 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 take a look at a few examples with detailed explanation the course of the decision 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 a cast 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 from 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.

\(\bullet\) A rational equation is an equation expressed as \[\dfrac(P(x))(Q(x))=0\] where \(P(x), \ Q(x)\) - polynomials (the sum of “xes” in various degrees, multiplied by various numbers).
The expression on the left side of the equation is called the rational expression.
The ODV (range of acceptable values) of a rational equation is all values \(x\) for which the denominator does NOT vanish, i.e. \(Q(x)\ne 0\) .
\(\bullet\) For example, equations \[\dfrac(x+2)(x-3)=0,\qquad \dfrac 2(x^2-1)=3, \qquad x^5-3x=2\] are rational equations.
In the first equation, the ODZ is all \(x\) such that \(x\ne 3\) (they write \(x\in (-\infty;3)\cup(3;+\infty)\)); in the second equation, these are all \(x\) , such that \(x\ne -1; x\ne 1\) (write \(x\in (-\infty;-1)\cup(-1;1)\cup(1;+\infty)\)); and in the third equation there are no restrictions on the ODZ, that is, the ODZ is all \(x\) (they write \(x\in\mathbb(R)\) ). \(\bullet\) Theorems:
1) The product of two factors is equal to zero if and only if one of them is equal to zero, while the other does not lose its meaning, therefore, the equation \(f(x)\cdot g(x)=0\) is equivalent to the system \[\begin(cases) \left[ \begin(gathered)\begin(aligned) &f(x)=0\\ &g(x)=0 \end(aligned) \end(gathered) \right.\\ \ text(ODV equations) \end(cases)\] 2) The fraction is equal to zero if and only if the numerator is equal to zero and the denominator is not equal to zero, therefore, the equation \(\dfrac(f(x))(g(x))=0\) is equivalent to the system of equations \[\begin(cases) f(x)=0\\ g(x)\ne 0 \end(cases)\]\(\bullet\) Let's look at some examples.

1) Solve the equation \(x+1=\dfrac 2x\) . Let's find the ODZ of this equation - this is \(x\ne 0\) (since \(x\) is in the denominator).
So, the ODZ can be written as follows: .
Let's transfer all the terms into one part and reduce to a common denominator: \[\dfrac((x+1)\cdot x)x-\dfrac 2x=0\quad\Leftrightarrow\quad \dfrac(x^2+x-2)x=0\quad\Leftrightarrow\quad \begin( cases) x^2+x-2=0\\x\ne 0\end(cases)\] The solution to the first equation of the system will be \(x=-2, x=1\) . We see that both roots are non-zero. Therefore, the answer is: \(x\in \(-2;1\)\) .

2) Solve the equation \(\left(\dfrac4x - 2\right)\cdot (x^2-x)=0\). Let us find the ODZ of this equation. We see that the only value \(x\) for which the left side does not make sense is \(x=0\) . So the OD can be written as follows: \(x\in (-\infty;0)\cup(0;+\infty)\).
Thus, this equation is equivalent to the system:

\[\begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x-2=0\\ &x^2-x=0 \end(aligned) \end(gathered) \right. \\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &\dfrac 4x=2\\ &x(x-1)= 0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered)\begin(aligned) &x =2\\ &x=1\\ &x=0 \end(aligned) \end(gathered) \right.\\ x\ne 0 \end(cases) \quad \Leftrightarrow \quad \left[ \begin(gathered) \begin(aligned) &x=2\\ &x=1 \end(aligned) \end(gathered) \right.\] Indeed, despite the fact that \(x=0\) is the root of the second factor, if you substitute \(x=0\) in the original equation, it will not make sense, because the expression \(\dfrac 40\) is not defined.
So the solution to this equation is \(x\in \(1;2\)\) .

3) Solve the equation \[\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1)\] In our equation \(4x^2-1\ne 0\) , whence \((2x-1)(2x+1)\ne 0\) , i.e. \(x\ne -\frac12; \frac12\) .
We transfer all the terms to the left side and reduce to a common denominator:

\(\dfrac(x^2+4x)(4x^2-1)=\dfrac(3-x-x^2)(4x^2-1) \quad \Leftrightarrow \quad \dfrac(x^2+4x- 3+x+x^2)(4x^2-1)=0\quad \Leftrightarrow \quad \dfrac(2x^2+5x-3)(4x^2-1)=0 \quad \Leftrightarrow\)

\(\Leftrightarrow \quad \begin(cases) 2x^2+5x-3=0\\ 4x^2-1\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) (2x-1 )(x+3)=0\\ (2x-1)(2x+1)\ne 0 \end(cases) \quad \Leftrightarrow \quad \begin(cases) \left[ \begin(gathered) \begin( aligned) &x=\dfrac12\\ &x=-3 \end(aligned)\end(gathered) \right.\\ x\ne \dfrac 12\\ x\ne -\dfrac 12 \end(cases) \quad \ Leftrightarrow \quad x=-3\)

Answer: \(x\in \(-3\)\) .

Comment. If the answer consists of a finite set of numbers, then they can be written separated by semicolons in curly braces, as shown in the previous examples.

Tasks that require solving rational equations are encountered in the Unified State Examination in mathematics every year, therefore, in preparation for passing the certification test, graduates should definitely repeat the theory on this topic on their own. To be able to cope with such tasks, graduates who pass both the basic and the profile level of the exam must necessarily. Having mastered the theory and dealt with practical exercises on the topic "Rational Equations", students will be able to solve problems with any number of actions and expect to receive competitive points at the end of the exam.

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Class 9.

Lesson topic:"Fractional Rational Equations"

Lesson type: combined.

Goals:

1. Educational: give a definition of "fractional-rational equations", show ways to solve such equations.

2. Developing: development of skills and abilities to solve examples with this type of equations, to find the roots of fractional rational equations.

3. Educators: to cultivate attention, attentiveness, activity, accuracy; respect for the mother.

Tasks:to interest students in the subject, to show the importance of the ability to solve various equations and problems.

Material and technical equipment:

Multimedia projector, screen, presentation for the lesson "Fractional rational equations"

Time: 45 minutes

Lesson plan.

Lesson stages	Teacher activity	Student activities
I. Organizing time. (1 min.)	Greets students, checks their readiness for the lesson.	Welcome teachers.
II. Presentation of the topic and objectives of the lesson. (2 minutes)	Informs the topic and purpose of the lesson.	Write down the topic in a notebook.
III. Repetition of the topic. (2 minutes)	Asks questions on the repetition of the topic.	They answer questions.
IV. Learning new material. (15 minutes.)	Demonstrates slides, accompanies the story. Listens, asks targeted questions in the role of an ordinary participant	They discuss the subject with the teacher and receive information, if necessary, set goals, plan the trajectory of work. Develop a plan of action, form tasks. They search for information, collect data and facts of history, primarily examine the information received, and solve intermediate tasks.
V. Fizkultminutka. (1 min.)	Performs physical education	Perform physical education
VI. Fixing the material. (20 minutes.)	Problem solving, offers questions for reinforcement.	Solve problems in notebooks, at the blackboard, ask questions to the teacher.
VIII. Summing up the lesson. (4min)	Evaluates student work.	Talk about what they learned in class. They take away jobs.

DURING THE CLASSES

I. Reflection at the beginning of the lesson(music; presentation about mother).

Check readiness for the lesson.

II. Message new topic, goals and objectives:

Teacher: Hello! Please look at each other and smile with all your heart.

I would like to start today's lesson with the words of M. Gorky:

slide 1
Flowers don't bloom without the sun

without love there is no happiness

without women there is no love,

without a mother there is neither a poet nor a hero.

All the pride of the world comes from mothers.
(M. Gorky)

Teacher:

- What could be more sacred than the name of the mother! …

A person who has not yet taken a single step on the ground and is only just starting to “babble”, hesitantly and diligently adds up “mother” in syllables and, feeling his luck, laughs, happy ...

When the baby screams for the first time

And his mother touches him gently,

Her love... Oh, how disturbing she is.

Anxious every day and hour.

Guys, Mother's Day is coming soon, so today's lesson I want to connect with this topic. In the past lessons, we have learned to decide, to find the roots various equations, today we will continue to get acquainted with one of the types of equations - these are fractional rational equations, find out the importance of equations, and remember how to solve problems using equations. We will try not to let our mother down, we will decide carefully and without being distracted, prepare for the GIA. The mother of each of you wants her child to be the best. So today we have a lesson on learning a new topic (slide 2).

III. Repetition of the topic.

1. Checking homework(slide 3).

No. 925 (a, b), No. 935 (a, b), No. 936.

2. Verbally repeat(slide 3 ,4,5,6 ).

Let's repeat:

What is the name of this equation? How many roots does this equation have?

IV . Learning new material.(slide 7).

Teacher: The equation y (x ) =0 called fractional rational equation, if expression y (x ) is an fractional(i.e. contains division by an expression with variables).

To solve a rational equation, it must be converted into a linear or quadratic equation, solve this equation and discard those roots that are not included in the ODZ (range of acceptable values) of the original rational equation.

Open the textbook on p.78 and read the rule. You already worked with this topic in 8th grade.

Algorithm for solving fractional rational equations: ( slide 8).

(Appendix 1)

Teacher: And now, together with me, let's solve a fractional-rational equation using the algorithm (slide 9).

VI . Independent work(slide 10).

… Your letter. your native lines.

Your last maternal order:

“The laws of life are wise and cruel.

Live. Work hard. Do not spoil your eyes with tears.

My love is with you always. Forever.

You love life. She's good.

Love people. And remember - in a person

what's important? High soul.

Let's try to make sure that we have a "high soul". And for this you need to respect and love your parents, of course, try to study and pass the state exam well. exams. Let's get ready for the certification.

Independent work. Self-control - 4 options. Checking your honesty. The work is done in notebooks. In the course of the work, students determine for themselves an algorithm for solving fractional rational equations. On each desk - a table - a reminder "Algorithm for solving fractional rational equations." Appendix 1.

Option 1.

V a r i a n t 2.

Option 3.

Option 4.

Answers:

I option:
,
(
;
).

II option:
(
;
)

III option:
(

)

IV option:
,
(
;
).

VII . Physical education minute(slide 11).

Teacher: And now the workout.

Turn to me. I'm making suggestions. If it is fair, you get up; if not, then you remain seated.

1) 5x = 7 has a single root.

2) 0x = 0 has no roots.
3) If D 0, then the quadratic equation has two roots.
4) If D
5) The number of roots is not more degree equations.

VIII . Consolidation and repetition of material.(slide 12).

Teacher. Men in front of their loved ones want to look only courageous, only strong, only unbending. Maybe that's what makes them men. And only in front of their own mother they are not afraid to expose their weaknesses and failures, to admit their mistakes and losses, because no matter how far they have gone in their age and development, before her they are gray-haired - all the same children. And she understands with her heart that the poor and offended, first of all, need a mother more than anyone. Today everyone will have good grades, so I think there will be no offended ones.

We solve the problem No. 942 from the textbook. (Algebra - Grade 9 / Yu.N. Makarychev) (slide 13).


1st car	x -20 km/h	h
2nd car	x km/h	h

Solve the example on the board.(slide 14).

№289(a)

VII . Summing up the lesson.

What new did you learn in the lesson?

What did you learn in the lesson?

2. Algorithm for solving fractional rational equations:

The teacher evaluates the work of students and assigns grades.

Teacher. Acquiring the traits of a symbol and fulfilling a huge social mission, the mother never lost her usual human features, remaining a hospitable hostess and an intelligent companion, a diligent worker and a natural songwriter, wide at the feast and courageous in grief, open in joy and restrained in sadness, and always kind, understanding and feminine! I really want your parents' dreams to come true, may you be worthy people (slide 15).

VIII . Homework . No. 943, No. 940 (a, b), No. 290 (slide 16).

Appendix 1.

Algorithm for solving fractional rational equations:

Find the allowable values of the fractions included in the equation.

Find the common denominator of the fractions in the equation.

Multiply both sides of the equation by a common denominator.

Solve the resulting equation.

Exclude roots that are not included in the allowed values of the fractions of the equation .