Identical transformations of expressions are one of the content lines of the school mathematics course. Identical transformations are widely used in solving equations, inequalities, systems of equations and inequalities. In addition, identical transformations of expressions contribute to the development of intelligence, flexibility and rationality of thinking.
The proposed materials are intended for 8th grade students and include the theoretical foundations of identical transformations of rational and irrational expressions, types of tasks for transforming such expressions and the text of the test.
1. Theoretical foundations of identity transformations
Expressions in algebra are records consisting of numbers and letters connected by action signs.
https://pandia.ru/text/80/197/images/image002_92.gif" width="77" height="21 src=">.gif" width="20" height="21 src="> – algebraic expressions.
Depending on the operations, rational and irrational expressions are distinguished.
Algebraic expressions are called rational if relative to the letters included in it A, b, With, ... no other operations are performed except addition, multiplication, subtraction, division and exponentiation.
Algebraic expressions containing operations of extracting the root of a variable or raising a variable to a rational power that is not an integer are called irrational with respect to this variable.
An identity transformation of a given expression is the replacement of one expression with another that is identically equal to it on a certain set.
The following theoretical facts underlie identical transformations of rational and irrational expressions.
1. Properties of degrees with an integer exponent:
, n ON; A 1=A;
, n ON, A¹0; A 0=1, A¹0;
, A¹0;
, A¹0;
, A¹0;
, A¹0, b¹0;
, A¹0, b¹0.
2. Abbreviated multiplication formulas:
Where A, b, With– any real numbers;
Where A¹0, X 1 and X 2 – roots of the equation 
.
3. The main property of fractions and actions on fractions:
, Where b¹0, With¹0;
; 
;
4. Definition of an arithmetic root and its properties:
; , b#0; https://pandia.ru/text/80/197/images/image026_24.gif" width="84" height="32">; ; 
,
Where A, b– non-negative numbers, n ON, n³2, m ON, m³2.
1. Types of Expression Conversion Exercises
There are various types of exercises on identity transformations of expressions. First type: The conversion that needs to be performed is explicitly specified.
For example.
1. Represent it as a polynomial.
When performing this transformation, we used the rules of multiplication and subtraction of polynomials, the formula for abbreviated multiplication and the reduction of similar terms.
2. Factor into: 
.
When performing the transformation, we used the rule of placing the common factor out of brackets and 2 abbreviated multiplication formulas.
3. Reduce the fraction:
.
When performing the transformation, we used the removal of the common factor from brackets, commutative and contractile laws, 2 abbreviated multiplication formulas, and operations on powers.
4. Remove the factor from under the root sign if A³0, b³0, With³0: https://pandia.ru/text/80/197/images/image036_17.gif" width="432" height="27">
We used the rules for actions on roots and the definition of the modulus of a number.
5. Eliminate irrationality in the denominator of a fraction. 
.
Second type exercises are exercises in which the main transformation that needs to be performed is clearly indicated. In such exercises, the requirement is usually formulated in one of the following forms: simplify the expression, calculate. When performing such exercises, it is necessary first of all to identify which and in what order transformations need to be performed so that the expression takes on a more compact form than the given one, or a numerical result is obtained.
For example
6. Simplify the expression:
Solution:
.
Used rules for operating algebraic fractions and abbreviated multiplication formulas.
7. Simplify the expression:
.
If A³0, b³0, A¹ b.
We used abbreviated multiplication formulas, rules for adding fractions and multiplying irrational expressions, the identity https://pandia.ru/text/80/197/images/image049_15.gif" width="203" height="29">.
We used the operation of selecting a complete square, the identity https://pandia.ru/text/80/197/images/image053_11.gif" width="132 height=21" height="21">, if .
Proof:
Since , then and or or or , i.e. .
We used the condition and formula for the sum of cubes.
It should be borne in mind that conditions connecting variables can also be specified in exercises of the first two types.
For example.
10. Find if .
This lesson will cover basic information about rational expressions and their transformations, as well as examples of transformations of rational expressions. This topic summarizes the topics we have studied so far. Transformations of rational expressions involve addition, subtraction, multiplication, division, exponentiation of algebraic fractions, reduction, factorization, etc. As part of the lesson, we will look at what a rational expression is, and also analyze examples of their transformation.
Subject:Algebraic fractions. Arithmetic operations on algebraic fractions
Lesson:Basic information about rational expressions and their transformations
Definition
Rational expression is an expression consisting of numbers, variables, arithmetic operations and the operation of exponentiation.
Let's look at an example of a rational expression:
Special cases of rational expressions:
1st degree: 
;
2. monomial: ;
3. fraction: .
Converting a rational expression is a simplification of a rational expression. The order of actions when transforming rational expressions: first there are operations in brackets, then multiplication (division) operations, and then addition (subtraction) operations.
Let's look at several examples of transforming rational expressions.
Example 1
Solution:
Let's solve this example step by step. The action in parentheses is executed first.
Answer:
Example 2
Solution:
Answer:
Example 3
Solution:
Answer: .
Note: Perhaps, when you saw this example, an idea arose: reduce the fraction before reducing it to a common denominator. Indeed, it is absolutely correct: first it is advisable to simplify the expression as much as possible, and then transform it. Let's try to solve this same example in the second way.
As you can see, the answer turned out to be absolutely similar, but the solution turned out to be somewhat simpler.
In this lesson we looked at rational expressions and their transformations, as well as several specific examples of these transformations.
Bibliography
1. Bashmakov M.I. Algebra 8th grade. - M.: Education, 2004.
2. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra 8. - 5th ed. - M.: Education, 2010.
Rational expressions and fractions are the cornerstone of the entire algebra course. Those who learn to work with such expressions, simplify them and factor them, will essentially be able to solve any problem, since transforming expressions is an integral part of any serious equation, inequality, or even word problem.
In this video tutorial we will look at how to correctly use abbreviated multiplication formulas to simplify rational expressions and fractions. Let's learn to see these formulas where, at first glance, there is nothing. At the same time, we will repeat such a simple technique as factoring a quadratic trinomial through a discriminant.
As you probably already guessed from the formulas behind me, today we will study abbreviated multiplication formulas, or, more precisely, not the formulas themselves, but their use to simplify and reduce complex rational expressions. But, before moving on to solving examples, let's take a closer look at these formulas or remember them:
- $((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right)$ — difference of squares;
- $((\left(a+b \right))^(2))=((a)^(2))+2ab+((b)^(2))$ is the square of the sum;
- $((\left(a-b \right))^(2))=((a)^(2))-2ab+((b)^(2))$ — squared difference;
- $((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b)^( 2)) \right)$ is the sum of cubes;
- $((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^(2) ) \right)$ is the difference of cubes.
I would also like to note that our school education system is structured in such a way that it is with the study of this topic, i.e. rational expressions, as well as roots, modules, all students have the same problem, which I will now explain.
The fact is that at the very beginning of studying abbreviated multiplication formulas and, accordingly, actions to reduce fractions (this is somewhere in the 8th grade), teachers say something like the following: “If something is not clear to you, then don’t worry, we will help you.” We will return to this topic more than once, in high school for sure. We'll look into this later." Well, then, at the turn of 9-10 grades, the same teachers explain to the same students who still don’t know how to solve rational fractions, something like this: “Where were you the previous two years? This was studied in algebra in 8th grade! What could be unclear here? It’s so obvious!”
However, such explanations do not make it any easier for ordinary students: they still had a mess in their heads, so right now we will look at two simple examples, on the basis of which we will see how to isolate these expressions in real problems, which will lead us to abbreviated multiplication formulas and how to then apply this to transform complex rational expressions.
Reducing simple rational fractions
Task No. 1
\[\frac(4x+3((y)^(2)))(9((y)^(4))-16((x)^(2)))\]
The first thing we need to learn is to identify exact squares and higher powers in the original expressions, on the basis of which we can then apply formulas. Let's get a look:
Let's rewrite our expression taking into account these facts:
\[\frac(4x+3((y)^(2)))(((\left(3((y)^(2)) \right))^(2))-((\left(4x \right))^(2)))=\frac(4x+3((y)^(2)))(\left(3((y)^(2))-4x \right)\left(3 ((y)^(2))+4x \right))=\frac(1)(3((y)^(2))-4x)\]
Answer: $\frac(1)(3((y)^(2))-4x)$.
Problem No. 2
Let's move on to the second task:
\[\frac(8)(((x)^(2))+5xy-6((y)^(2)))\]
There is nothing to simplify here, because the numerator contains a constant, but I proposed this problem precisely so that you learn how to factor polynomials containing two variables. If instead we had the polynomial below, how would we expand it?
\[((x)^(2))+5x-6=\left(x-... \right)\left(x-... \right)\]
Let's solve the equation and find the $x$ that we can put in place of the dots:
\[((x)^(2))+5x-6=0\]
\[((x)_(1))=\frac(-5+7)(2)=\frac(2)(2)=1\]
\[((x)_(2))=\frac(-5-7)(2)=\frac(-12)(2)=-6\]
We can rewrite the trinomial as follows:
\[((x)^(2))+5xy-6((y)^(2))=\left(x-1 \right)\left(x+6 \right)\]
We learned how to work with a quadratic trinomial - that's why we needed to record this video lesson. But what if, in addition to $x$ and a constant, there is also $y$? Let's consider them as another element of the coefficients, i.e. Let's rewrite our expression as follows:
\[((x)^(2))+5y\cdot x-6((y)^(2))\]
\[((x)_(1))=\frac(-5y+7y)(2)=y\]
\[((x)_(2))=\frac(-5y-7y)(2)=\frac(-12y)(2)=-6y\]
Let us write the expansion of our square construction:
\[\left(x-y \right)\left(x+6y \right)\]
So, if we return to the original expression and rewrite it taking into account the changes, we get the following:
\[\frac(8)(\left(x-y \right)\left(x+6y \right))\]
What does such a record give us? Nothing, because it cannot be reduced, it is not multiplied or divided by anything. However, as soon as this fraction turns out to be an integral part of a more complex expression, such an expansion will come in handy. Therefore, as soon as you see a quadratic trinomial (it does not matter whether it is burdened with additional parameters or not), always try to factor it.
Nuances of the solution
Remember the basic rules for converting rational expressions:
- All denominators and numerators must be factored either through abbreviated multiplication formulas or through a discriminant.
- You need to work according to the following algorithm: when we look and try to isolate the formula for abbreviated multiplication, then, first of all, we try to convert everything to the highest possible degree. After this, we take the overall degree out of the bracket.
- Very often you will encounter expressions with a parameter: other variables will appear as coefficients. We find them using the quadratic expansion formula.
So, once you see rational fractions, the first thing to do is factor both the numerator and denominator into linear expressions, using the abbreviated multiplication or discriminant formulas.
Let's look at a couple of these rational expressions and try to factor them.
Solving more complex examples
Task No. 1
\[\frac(4((x)^(2))-6xy+9((y)^(2)))(2x-3y)\cdot \frac(9((y)^(2))- 4((x)^(2)))(8((x)^(3))+27((y)^(3)))\]
We rewrite and try to decompose each term:
Let's rewrite our entire rational expression taking into account these facts:
\[\frac(((\left(2x \right))^(2))-2x\cdot 3y+((\left(3y \right))^(2)))(2x-3y)\cdot \frac (((\left(3y \right))^(2))-((\left(2x \right))^(2)))(((\left(2x \right))^(3))+ ((\left(3y \right))^(3)))=\]
\[=\frac(((\left(2x \right))^(2))-2x\cdot 3y+((\left(3y \right))^(2)))(2x-3y)\cdot \ frac(\left(3y-2x \right)\left(3y+2x \right))(\left(2x+3y \right)\left(((\left(2x \right))^(2))- 2x\cdot 3y+((\left(3y \right))^(2)) \right))=-1\]
Answer: $-1$.
Problem No. 2
\[\frac(3-6x)(2((x)^(2))+4x+8)\cdot \frac(2x+1)(((x)^(2))+4-4x)\ cdot \frac(8-((x)^(3)))(4((x)^(2))-1)\]
Let's look at all the fractions.
\[((x)^(2))+4-4x=((x)^(2))-4x+2=((x)^(2))-2\cdot 2x+((2)^( 2))=((\left(x-2 \right))^(2))\]
Let's rewrite the entire structure taking into account the changes:
\[\frac(3\left(1-2x \right))(2\left(((x)^(2))+2x+((2)^(2)) \right))\cdot \frac( 2x+1)(((\left(x-2 \right))^(2)))\cdot \frac(\left(2-x \right)\left(((2)^(2))+ 2x+((x)^(2)) \right))(\left(2x-1 \right)\left(2x+1 \right))=\]
\[=\frac(3\cdot \left(-1 \right))(2\cdot \left(x-2 \right)\cdot \left(-1 \right))=\frac(3)(2 \left(x-2 \right))\]
Answer: $\frac(3)(2\left(x-2 \right))$.
Nuances of the solution
So what we just learned:
- Not every square trinomial can be factorized; in particular, this applies to the incomplete square of the sum or difference, which are very often found as parts of sum or difference cubes.
- Constants, i.e. ordinary numbers that do not have variables can also act as active elements in the expansion process. Firstly, they can be taken out of brackets, and secondly, the constants themselves can be represented in the form of powers.
- Very often, after factoring all the elements, opposite constructions arise. These fractions must be reduced extremely carefully, because when crossing them out either above or below, an additional factor $-1$ appears - this is precisely a consequence of the fact that they are opposites.
Solving complex problems
\[\frac(27((a)^(3))-64((b)^(3)))(((b)^(2))-4):\frac(9((a)^ (2))+12ab+16((b)^(2)))(((b)^(2))+4b+4)\]
Let's consider each term separately.
First fraction:
\[((\left(3a \right))^(3))-((\left(4b \right))^(3))=\left(3a-4b \right)\left(((\left (3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2)) \right)\]
\[((b)^(2))-((2)^(2))=\left(b-2 \right)\left(b+2 \right)\]
We can rewrite the entire numerator of the second fraction as follows:
\[((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2))\]
Now let's look at the denominator:
\[((b)^(2))+4b+4=((b)^(2))+2\cdot 2b+((2)^(2))=((\left(b+2 \right ))^(2))\]
Let's rewrite the entire rational expression taking into account the above facts:
\[\frac(\left(3a-4b \right)\left(((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2 )) \right))(\left(b-2 \right)\left(b+2 \right))\cdot \frac(((\left(b+2 \right))^(2)))( ((\left(3a \right))^(2))+3a\cdot 4b+((\left(4b \right))^(2)))=\]
\[=\frac(\left(3a-4b \right)\left(b+2 \right))(\left(b-2 \right))\]
Answer: $\frac(\left(3a-4b \right)\left(b+2 \right))(\left(b-2 \right))$.
Nuances of the solution
As we have seen once again, incomplete squares of the sum or incomplete squares of the difference, which are often found in real rational expressions, however, do not be afraid of them, because after transforming each element they are almost always canceled. In addition, in no case should you be afraid of large constructions in the final answer - it is quite possible that this is not your mistake (especially if everything is factorized), but the author intended such an answer.
In conclusion, I would like to look at another complex example, which no longer directly relates to rational fractions, but it contains everything that awaits you on real tests and exams, namely: factorization, reduction to a common denominator, reduction of similar terms. This is exactly what we will do now.
Solving a complex problem of simplifying and transforming rational expressions
\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]
First, let's look at and open the first bracket: in it we see three separate fractions with different denominators, so the first thing we need to do is bring all three fractions to a common denominator, and to do this, each of them should be factored:
\[((x)^(2))+2x+4=((x)^(2))+2\cdot x+((2)^(2))\]
\[((x)^(2))-8=((x)^(3))-((2)^(2))=\left(x-2 \right)\left(((x) ^(2))+2x+((2)^(2)) \right)\]
Let's rewrite our entire construction as follows:
\[\frac(x)(((x)^(2))+2x+((2)^(2)))+\frac(((x)^(2))+8)(\left(x -2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))-\frac(1)(x-2)=\]
\[=\frac(x\left(x-2 \right)+((x)^(3))+8-\left(((x)^(2))+2x+((2)^(2 )) \right))(\left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\]
\[=\frac(((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\frac(((x)^(2))-4x-4)(\ left(x-2 \right)\left(((x)^(2))+2x+((2)^(2)) \right))=\]
\[=\frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+(( 2)^(2)) \right))=\frac(x-2)(((x)^(2))+2x+4)\]
This is the result of the calculations from the first bracket.
Let's deal with the second bracket:
\[((x)^(2))-4=((x)^(2))-((2)^(2))=\left(x-2 \right)\left(x+2 \ right)\]
Let's rewrite the second bracket taking into account the changes:
\[\frac(((x)^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)=\frac( ((x)^(2))+2\left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^ (2))+2x+4)(\left(x-2 \right)\left(x+2 \right))\]
Now let's write down the entire original construction:
\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]
Answer: $\frac(1)(x+2)$.
Nuances of the solution
As you can see, the answer turned out to be quite reasonable. However, please note: very often during such large-scale calculations, when the only variable appears only in the denominator, students forget that this is the denominator and it should be at the bottom of the fraction and write this expression in the numerator - this is a gross mistake.
In addition, I would like to draw your special attention to how such tasks are formalized. In any complex calculations, all steps are performed one by one: first we count the first bracket separately, then the second one separately, and only at the end do we combine all the parts and calculate the result. In this way, we insure ourselves against stupid mistakes, carefully write down all the calculations and at the same time do not waste any extra time, as it might seem at first glance.
The article talks about the transformation of rational expressions. Let's consider the types of rational expressions, their transformations, groupings, and bracketing the common factor. Let's learn to represent fractional rational expressions in the form of rational fractions.
Yandex.RTB R-A-339285-1
Definition and examples of rational expressions
Definition 1Expressions that are made up of numbers, variables, parentheses, powers with the operations of addition, subtraction, multiplication, division with the presence of a fraction line are called rational expressions.
For example, we have that 5, 2 3 x - 5, - 3 a b 3 - 1 c 2 + 4 a 2 + b 2 1 + a: (1 - b) , (x + 1) (y - 2) x 5 - 5 · x · y · 2 - 1 11 · x 3 .
That is, these are expressions that are not divided into expressions with variables. The study of rational expressions begins in grade 8, where they are called fractional rational expressions. Particular attention is paid to fractions in the numerator, which are transformed using transformation rules.
This allows us to proceed to the transformation of rational fractions of arbitrary form. Such an expression can be considered as an expression with the presence of rational fractions and integer expressions with action signs.
Main types of transformations of rational expressions
Rational expressions are used to perform identical transformations, groupings, bringing similar ones, and performing other operations with numbers. The purpose of such expressions is simplification.
Example 1
Convert the rational expression 3 · x x · y - 1 - 2 · x x · y - 1 .
Solution
It can be seen that such a rational expression is the difference between 3 x x y - 1 and 2 x x y - 1. We notice that their denominator is identical. This means that the reduction of similar terms will take the form
3 x x y - 1 - 2 x x y - 1 = x x y - 1 3 - 2 = x x y - 1
Answer: 3 · x x · y - 1 - 2 · x x · y - 1 = x x · y - 1 .
Example 2
Convert 2 x y 4 (- 4) x 2: (3 x - x) .
Solution
Initially, we perform the actions in brackets 3 · x − x = 2 · x. We present this expression in the form 2 · x · y 4 · (- 4) · x 2: (3 · x - x) = 2 · x · y 4 · (- 4) · x 2: 2 · x. We arrive at an expression that contains operations with one step, that is, it has addition and subtraction.
We get rid of parentheses by using the division property. Then we get that 2 · x · y 4 · (- 4) · x 2: 2 · x = 2 · x · y 4 · (- 4) · x 2: 2: x.
We group numerical factors with the variable x, after which we can perform operations with powers. We get that
2 x y 4 (- 4) x 2: 2: x = (2 (- 4) : 2) (x x 2: x) y 4 = - 4 x 2 y 4
Answer: 2 x y 4 (- 4) x 2: (3 x - x) = - 4 x 2 y 4.
Example 3
Transform an expression of the form x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 .
Solution
First, we transform the numerator and denominator. Then we get an expression of the form (x · (x + 3) - (3 · x + 1)) : 1 2 · x · 4 + 2 , and the actions in parentheses are done first. In the numerator, operations are performed and factors are grouped. Then we get an expression of the form x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 = x 2 + 3 · x - 3 · x - 1 1 2 · 4 · x + 2 = x 2 - 1 2 · x + 2 .
We transform the difference of squares formula in the numerator, then we get that
x 2 - 1 2 x + 2 = (x - 1) (x + 1) 2 (x + 1) = x - 1 2
Answer: x · (x + 3) - (3 · x + 1) 1 2 · x · 4 + 2 = x - 1 2 .
Rational fraction representation
Algebraic fractions are most often simplified when solved. Each rational is brought to this in different ways. It is necessary to perform all the necessary operations with polynomials so that the rational expression can ultimately give a rational fraction.
Example 4
Present as a rational fraction a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a.
Solution
This expression can be represented as a 2 - 25 a + 3 · 1 a 2 + 5 · a. Multiplication is performed primarily according to the rules.
We should start with multiplication, then we get that
a 2 - 25 a + 3 1 a 2 + 5 a = a - 5 (a + 5) a + 3 1 a (a + 5) = a - 5 (a + 5) 1 ( a + 3) a (a + 5) = a - 5 (a + 3) a
We present the obtained result with the original one. We get that
a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a = a + 5 a · a - 3 - a - 5 a + 3 · a
Now let's do the subtraction:
a + 5 a · a - 3 - a - 5 a + 3 · a = a + 5 · a + 3 a · (a - 3) · (a + 3) - (a - 5) · (a - 3) (a + 3) a (a - 3) = = a + 5 a + 3 - (a - 5) (a - 3) a (a - 3) (a + 3) = a 2 + 3 a + 5 a + 15 - (a 2 - 3 a - 5 a + 15) a (a - 3) (a + 3) = = 16 a a (a - 3) (a + 3) = 16 a - 3 (a + 3) = 16 a 2 - 9
After which it is obvious that the original expression will take the form 16 a 2 - 9.
Answer: a + 5 a · (a - 3) - a 2 - 25 a + 3 · 1 a 2 + 5 · a = 16 a 2 - 9 .
Example 5
Express x x + 1 + 1 2 · x - 1 1 + x as a rational fraction.
Solution
The given expression is written as a fraction, the numerator of which has x x + 1 + 1, and the denominator 2 x - 1 1 + x. It is necessary to make transformations x x + 1 + 1 . To do this you need to add a fraction and a number. We get that x x + 1 + 1 = x x + 1 + 1 1 = x x + 1 + 1 · (x + 1) 1 · (x + 1) = x x + 1 + x + 1 x + 1 = x + x + 1 x + 1 = 2 x + 1 x + 1
It follows that x x + 1 + 1 2 x - 1 1 + x = 2 x + 1 x + 1 2 x - 1 1 + x
The resulting fraction can be written as 2 x + 1 x + 1: 2 x - 1 1 + x.
After division we arrive at a rational fraction of the form
2 x + 1 x + 1: 2 x - 1 1 + x = 2 x + 1 x + 1 1 + x 2 x - 1 = 2 x + 1 (1 + x) (x + 1) (2 x - 1) = 2 x + 1 2 x - 1
You can solve this differently.
Instead of dividing by 2 x - 1 1 + x, we multiply by its inverse 1 + x 2 x - 1. Let us apply the distribution property and find that
x x + 1 + 1 2 x - 1 1 + x = x x + 1 + 1: 2 x - 1 1 + x = x x + 1 + 1 1 + x 2 x - 1 = = x x + 1 1 + x 2 x - 1 + 1 1 + x 2 x - 1 = x 1 + x (x + 1) 2 x - 1 + 1 + x 2 x - 1 = = x 2 x - 1 + 1 + x 2 x - 1 = x + 1 + x 2 x - 1 = 2 x + 1 2 x - 1
Answer: x x + 1 + 1 2 · x - 1 1 + x = 2 · x + 1 2 · x - 1 .
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