How to solve x+1 x-1 + x-2 x+2 =4- 2x+3 x-2 ;x ne1,-2,2 ?

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Welcome to my article How to solve x+1 x-1 + x-2 x+2 =4- 2x+3 x-2 ;x ne1,-2,2 ?
This article has been taken from the simplification lesson, in this article we have been told to simplify with the action of addition, subtraction, multiplication, division and fractions.
For complete information on how to solve this x+1 x-1 + x-2 x+2 =4- 2x+3 x-2 ;x ne1,-2,2 , read and understand it carefully.

First of all we should write the article on the page of the notebook.

x+1 x-1 + x-2 x+2 =4- 2x+3 x-2 ;x ne1,-2,2 ?

We should use these methods while solving this type of question,
Let’s solve it by placing it up and down,

\displaystyle \frac{{x+1}}{{x-1}}+\frac{{x-2}}{{x+2}}=4-\frac{{2x+3}}{{x-2}}

While solving these types of questions, we find the LCM of the denominator of their terms.
Then after that solve like this.

\displaystyle \frac{{(x+1)(x-2)+(x-2)(x-1)}}{{(x-1)(x+2)}}=\frac{{4(x-2)-2x+3}}{{x-2}}

\displaystyle \frac{{{{x}^{2}}+2x+x+2+{{x}^{2}}-x-2x+2}}{{{{x}^{2}}+2x-x-2}}=\frac{{4x-8-2x-3}}{{x-2}}

\displaystyle \frac{{{{x}^{2}}+{{x}^{2}}+2x+x-x-2x+2+2}}{{{{x}^{2}}+2x-x-2}}=\frac{{4x-2x-3-8}}{{x-2}}

\displaystyle \frac{{2{{x}^{2}}+3x-3x+4}}{{{{x}^{2}}+x-2}}=\frac{{4x-2x-3-8}}{{x-2}}

\displaystyle \frac{{2{{x}^{2}}+4}}{{{{x}^{2}}+x-2}}=\frac{{2x-11}}{{x-2}}

Then multiply these terms diagonally.

In doing so,

\displaystyle (2{{x}^{2}}+4)(x+2)=(2x-11)({{x}^{2}}+x-2)

\displaystyle 2x.{{x}^{2}}-4{{x}^{2}}+4x-8=2x.{{x}^{2}}+2{{x}^{2}}-4x-11{{x}^{2}}-11x+22

\displaystyle 2x.{{x}^{2}}-2x.{{x}^{2}}-4{{x}^{2}}-2{{x}^{2}}+11{{x}^{2}}+4x+4x+11x-8-22=0

\displaystyle -6{{x}^{2}}+11{{x}^{2}}+8x+11x-8-22=0

\displaystyle 5{{x}^{2}}+19x-8-22=0

\displaystyle 5{{x}^{2}}+19x-30=0

\displaystyle 5{{x}^{2}}+25x-6x-30=0

\displaystyle 5x(x+5)-6(x+5)=0

\displaystyle (x+5)(5x-6)=0

If
x +5=0
So ,
If ,
5x -6 = 0
So,