The least common denominator is used to simplify this equation. This method is used when you cannot write the given equation with one rational expression on each side of the equation (and use the cross multiplication method). This method is used when you are given a rational equation with 3 or more fractions (in the case of two fractions, cross multiplication is better).
Find the least common denominator of fractions (or least common multiple). NOZ is the smallest number that is evenly divisible by each denominator.
- Sometimes NOZ is an obvious number. For example, if the equation is given: x/3 + 1/2 = (3x + 1)/6, then it is obvious that the least common multiple of the numbers 3, 2 and 6 will be 6.
- If the NOD is not obvious, write down the multiples of the largest denominator and find among them one that is a multiple of the other denominators as well. You can often find the NOD by simply multiplying two denominators together. For example, if the equation x/8 + 2/6 = (x - 3)/9 is given, then NOZ = 8*9 = 72.
- If one or more denominators contain a variable, then the process is somewhat more complicated (but not impossible). In this case, the NOZ is an expression (containing a variable) that is divisible by each denominator. For example, in the equation 5/(x-1) = 1/x + 2/(3x) NOZ = 3x(x-1), because this expression is divisible by each denominator: 3x(x-1)/(x-1 ) = 3x; 3x(x-1)/3x = (x-1); 3x(x-1)/x = 3(x-1).
Multiply both the numerator and denominator of each fraction by a number equal to the result of dividing the NOZ by the corresponding denominator of each fraction. Since you're multiplying both the numerator and denominator by the same number, you're effectively multiplying a fraction by 1 (for example, 2/2 = 1 or 3/3 = 1).
- So in our example, multiply x/3 by 2/2 to get 2x/6, and multiply 1/2 by 3/3 to get 3/6 (3x + 1/6 does not need to be multiplied because it the denominator is 6).
- Proceed similarly when the variable is in the denominator. In our second example NOZ = 3x(x-1), so 5/(x-1) times (3x)/(3x) is 5(3x)/(3x)(x-1); 1/x times 3(x-1)/3(x-1) to get 3(x-1)/3x(x-1); 2/(3x) multiply by (x-1)/(x-1) and you get 2(x-1)/3x(x-1).
Find x. Now that you've reduced the fractions to a common denominator, you can get rid of the denominator. To do this, multiply each side of the equation by a common denominator. Then solve the resulting equation, that is, find "x". To do this, isolate the variable on one side of the equation.
- In our example: 2x/6 + 3/6 = (3x +1)/6. You can add 2 fractions with the same denominator, so write the equation as: (2x+3)/6=(3x+1)/6. Multiply both sides of the equation by 6 and get rid of the denominators: 2x+3 = 3x +1. Solve and get x = 2.
- In our second example (with a variable in the denominator), the equation looks like (after reduction to a common denominator): 5(3x)/(3x)(x-1) = 3(x-1)/3x(x-1) + 2 (x-1)/3x(x-1). By multiplying both sides of the equation by NOZ, you get rid of the denominator and get: 5(3x) = 3(x-1) + 2(x-1), or 15x = 3x - 3 + 2x -2, or 15x = x - 5 Solve and get: x = -5/14.
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased. In the 5th grade, students in mathematics study quite a lot of new topics, one of which will be fractional equations. For many, this is a rather complicated topic that parents should help their children understand, and if parents have forgotten math, they can always use online programs that solve equations. So, using an example, you can quickly understand the algorithm for solving equations with fractions and help your child.
Below, for clarity, we will solve a simple fractional linear equation of the following form:
\[\frac(x-2)(3) - \frac(3x)(2)=5\]
To solve this kind of equation, it is necessary to determine the NOZ and multiply the left and right sides of the equation by it:
\[\frac(x-2)(3) - \frac(3x)(2)=5\]
This will give us a simple linear equation because the common denominator as well as the denominator of each fractional term cancels out:
Let's move the terms from the unknown to the left side:
Let's divide the left and right parts by -7:
From the result obtained, an integer part can be distinguished, which will be the final result of solving this fractional equation:
Where can I solve the equation with fractions online?
You can solve the equation on our website https: // site. Free online solver will allow you to solve an online equation of any complexity in seconds. All you have to do is just enter your data into the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you have any questions, you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
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.
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 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 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.
Solution.
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 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) .
Solution.
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).
Solution.
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 entire 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 is equal to 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.
Solution.
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.
Solution.
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.
Solution.
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 located 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.
Solution.
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.
Solution.
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.
Solution.
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.
Solution.
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.
Solution.
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.
"Solution of fractional rational equations"
Lesson Objectives:
Tutorial:
- formation of the concept of fractional rational equations; to consider various ways of solving fractional rational equations; consider an algorithm for solving fractional rational equations, including the condition that the fraction is equal to zero; to teach the solution of fractional rational equations according to the algorithm; checking the level of assimilation of the topic by conducting test work.
Developing:
- development of the ability to correctly operate with the acquired knowledge, to think logically; development of intellectual skills and mental operations - analysis, synthesis, comparison and generalization; development of initiative, the ability to make decisions, not to stop there; development of critical thinking; development of research skills.
Nurturing:
- education of cognitive interest in the subject; education of independence in solving educational problems; education of will and perseverance to achieve the final results.
Lesson type: lesson - explanation of new material.
During the classes
1. Organizational moment.
Hello guys! Equations are written on the blackboard, look at them carefully. Can you solve all of these equations? Which ones are not and why?
Equations in which the left and right sides are fractional rational expressions are called fractional rational equations. What do you think we will study today in the lesson? Formulate the topic of the lesson. So, we open notebooks and write down the topic of the lesson “Solution of fractional rational equations”.
2. Actualization of knowledge. Frontal survey, oral work with the class.
And now we will repeat the main theoretical material that we need to study a new topic. Please answer the following questions:
1. What is an equation? ( Equality with a variable or variables.)
2. What is Equation #1 called? ( Linear.) Method for solving linear equations. ( Move everything with the unknown to the left side of the equation, all numbers to the right. Bring like terms. Find the unknown multiplier).
3. What is Equation #3 called? ( Square.) Methods for solving quadratic equations. ( Selection of the full square, by formulas, using the Vieta theorem and its consequences.)
4. What is a proportion? ( Equality of two relations.) The main property of proportion. ( If the proportion is true, then the product of its extreme terms is equal to the product of the middle terms.)
5. What properties are used in solving equations? ( 1. If in the equation we transfer the term from one part to another, changing its sign, then we get an equation equivalent to the given one. 2. If both parts of the equation are multiplied or divided by the same non-zero number, then an equation will be obtained that is equivalent to the given.)
6. When is a fraction equal to zero? ( A fraction is zero when the numerator is zero and the denominator is non-zero.)
3. Explanation of new material.
Solve equation No. 2 in notebooks and on the board.
Answer: 10.
What fractional rational equation can you try to solve using the basic property of proportion? (No. 5).
(x-2)(x-4) = (x+2)(x+3)
x2-4x-2x+8 = x2+3x+2x+6
x2-6x-x2-5x = 6-8
Solve equation No. 4 in notebooks and on the board.
Answer: 1,5.
What fractional rational equation can you try to solve by multiplying both sides of the equation by the denominator? (No. 6).
D=1>0, x1=3, x2=4.
Answer: 3;4.
Now try to solve equation #7 in one of the ways.
(x2-2x-5)x(x-5)=x(x-5)(x+5) |
|
||
(x2-2x-5)x(x-5)-x(x-5)(x+5)=0 | |||
x(x-5)(x2-2x-5-(x+5))=0 | x2-2x-5-x-5=0 |
||
x(x-5)(x2-3x-10)=0 | |||
x=0 x-5=0 x2-3x-10=0 | |||
x1=0 x2=5 D=49 | |||
Answer: 0;5;-2. | Answer: 5;-2. |
Explain why this happened? Why are there three roots in one case and two in the other? What numbers are the roots of this fractional rational equation?
Until now, students have not met the concept of an extraneous root, it is really very difficult for them to understand why this happened. If no one in the class can give a clear explanation of this situation, then the teacher asks leading questions.
- How do equations No. 2 and 4 differ from equations No. 5,6,7? ( In equations No. 2 and 4 in the denominator of the number, No. 5-7 - expressions with a variable.) What is the root of the equation? ( The value of the variable at which the equation becomes a true equality.) How to find out if the number is the root of the equation? ( Make a check.)
When doing a test, some students notice that they have to divide by zero. They conclude that the numbers 0 and 5 are not the roots of this equation. The question arises: is there a way to solve fractional rational equations that eliminates this error? Yes, this method is based on the condition that the fraction is equal to zero.
x2-3x-10=0, D=49, x1=5, x2=-2.
If x=5, then x(x-5)=0, so 5 is an extraneous root.
If x=-2, then x(x-5)≠0.
Answer: -2.
Let's try to formulate an algorithm for solving fractional rational equations in this way. Children themselves formulate the algorithm.
Algorithm for solving fractional rational equations:
1. Move everything to the left side.
2. Bring fractions to a common denominator.
3. Make a system: the fraction is equal to zero when the numerator is equal to zero, and the denominator is not equal to zero.
4. Solve the equation.
5. Check the inequality to exclude extraneous roots.
6. Write down the answer.
Discussion: how to formalize the solution if the basic property of proportion is used and the multiplication of both sides of the equation by a common denominator. (Supplement the solution: exclude from its roots those that turn the common denominator to zero).
4. Primary comprehension of new material.
Work in pairs. Students choose how to solve the equation on their own, depending on the type of equation. Tasks from the textbook "Algebra 8", 2007: No. 000 (b, c, i); No. 000(a, e, g). The teacher controls the performance of the task, answers the questions that have arisen, and provides assistance to poorly performing students. Self-test: Answers are written on the board.
b) 2 is an extraneous root. Answer:3.
c) 2 is an extraneous root. Answer: 1.5.
a) Answer: -12.5.
g) Answer: 1; 1.5.
5. Statement of homework.
2. Learn the algorithm for solving fractional rational equations.
3. Solve in notebooks No. 000 (a, d, e); No. 000(g, h).
4. Try to solve No. 000(a) (optional).
6. Fulfillment of the control task on the studied topic.
The work is done on sheets.
Job example:
A) Which of the equations are fractional rational?
B) A fraction is zero when the numerator is ______________________ and the denominator is _______________________.
Q) Is the number -3 the root of Equation #6?
D) Solve equation No. 7.
Task evaluation criteria:
- "5" is given if the student completed more than 90% of the task correctly. "4" - 75% -89% "3" - 50% -74% "2" is given to the student who completed less than 50% of the task. Grade 2 is not put in the journal, 3 is optional.
7. Reflection.
On the leaflets with independent work, put:
- 1 - if the lesson was interesting and understandable to you; 2 - interesting, but not clear; 3 - not interesting, but understandable; 4 - not interesting, not clear.
8. Summing up the lesson.
So, today in the lesson we got acquainted with fractional rational equations, learned how to solve these equations in various ways, tested our knowledge with the help of educational independent work. You will learn the results of independent work in the next lesson, at home you will have the opportunity to consolidate the knowledge gained.
What method of solving fractional rational equations, in your opinion, is easier, more accessible, more rational? Regardless of the method of solving fractional rational equations, what should not be forgotten? What is the "cunning" of fractional rational equations?
Thank you all, the lesson is over.
