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