#andy

\[\begin{array}{l}
\dfrac{{x + 1}}{{2x - 2}} - \dfrac{{2x}}{{{x^2} - 1}}\\
DKXD:\left\{ \begin{array}{l}
x \ne 1\\
x \ne  - 1
\end{array} \right.\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} - \dfrac{{2x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - 4x}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1 - 4x}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{x - 1}}{{2\left( {x + 1} \right)}}\\
 = \dfrac{{x - 1}}{{2x + 2}}
\end{array}\]

Đáp án:

`(x-1)/(2x+2)` 

Giải thích các bước giải:

`(x+1)/(2x-2)-(2x)/(x^2-1)` `(x \ne ±1)`

`=(x+1)/(2(x-1))-(2x)/((x-1)(x+1))`

`=((x+1)^2-4x)/(2(x-1)(x+1))`

`=(x^2+2x+1-4x)/(2(x-1)(x+1))`

`=((x-1)^2)/(2(x-1)(x+1))`

`=(x-1)/(2(x+1))`

`=(x-1)/(2x+2)`