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