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