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