Đáp án: $P = \dfrac{2}{{x - 1}}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x\# 0;x\# 1;x\# - 1\\
P = \left( {\dfrac{{{x^2} + 1}}{{{x^2} - 1}} - \dfrac{1}{{x - 1}}} \right)\left( {\dfrac{4}{{x - 1}} - \dfrac{2}{x}} \right)\\
= \left( {\dfrac{{{x^2} + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{1}{{x - 1}}} \right).\dfrac{{4x - 2\left( {x - 1} \right)}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{{x^2} + 1 - \left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{2x + 2}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{{x^2} - x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{2\left( {x + 1} \right)}}{{{x^2} - x}}\\
= \dfrac{2}{{x - 1}}
\end{array}$
