Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
P = \left( {\sqrt x - \dfrac{{x + 2}}{{\sqrt x + 1}}} \right):\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{{x - 4}}{{1 - x}}} \right)\\
= \left( {\dfrac{{x + \sqrt x - x - 2}}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - \sqrt x + x - 4}}\\
= \dfrac{{\sqrt x - 2}}{1}.\dfrac{{\sqrt x - 1}}{{2x - \sqrt x - 4}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{2x - \sqrt x - 4}}
\end{array}$
