Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 1\\
A = \left( {\dfrac{{\sqrt x - 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \left( {\dfrac{{\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}}} \right).\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{x - \sqrt x - 2 - \left( {x + \sqrt x - 2} \right)}}{1}.\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{ - 2\sqrt x }}{1}.\dfrac{{\sqrt x - 1}}{2}\\
= - \sqrt x .\left( {\sqrt x - 1} \right)\\
= \sqrt x - x\\
b)x - 3\sqrt x + 2 = 0\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {ktm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.\\
Thay\,x = 4\,vào\\
\Rightarrow A = \sqrt x - x = 2 - 4 = - 2
\end{array}$
Vậy A=-2
