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