# How to solve a=(1/1-x+2/x+1-5-x/1-x^2) 1-2x/x^2-1 ?

Welcome to my article How to solve a=(1/1-x+2/x+1-5-x/1-x^2) 1-2x/x^2-1 ?. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve a=(1/1-x+2/x+1-5-x/1-x^2) 1-2x/x^2-1 ?, read and understand it carefully till the end.

Let us know how to solve this questionHow to solve a=(1/1-x+2/x+1-5-x/1-x^2) 1-2x/x^2-1 ?

First write the question on the page of the notebook.

## How to solve a=(1/1-x+2/x+1-5-x/1-x^2) 1-2x/x^2-1 ?

By writing this question correctly in this way,

\displaystyle a=\left( {\frac{1}{1}-x+\frac{2}{x}+1-5+\frac{{-x}}{1}-1{{x}^{2}}} \right)\times 1+\frac{{-2x}}{{{{x}^{2}}}}-1 ,

Let’s solve it –

\displaystyle a=\left( {1-x+\frac{2}{x}+1-5-\frac{x}{1}-{{x}^{2}}} \right)\times 1-\frac{2}{x}-1 ,

\displaystyle a=\left( {1-x+\frac{2}{x}-4-x-{{x}^{2}}} \right)\times 1-\frac{2}{x}-1 ,

\displaystyle a=1-x+\frac{2}{x}-4-x-{{x}^{2}}-\frac{2}{x}-1 ,

\displaystyle a=1-2x+\frac{2}{x}-4-{{x}^{2}}-\frac{2}{x}-1 ,

\displaystyle a=1-2x-4-{{x}^{2}}-1 ,

\displaystyle a=-2x-4-{{x}^{2}} ,

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

or,

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

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

\displaystyle a=-[x(x+4)-2(x-2)] ,