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