Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{x\sqrt x - \sqrt x + x - 1}}} \right):\left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{2}{{x - 1}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{\sqrt x \left( {x - 1} \right) + x - 1}}} \right):\left( {\dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}} \right).\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= \dfrac{{x - 1 - 2\sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {x - 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{x - 2\sqrt x + 1}}{{x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}
\end{array}$
