Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x\# 1;x\# 4\\
A = \left( {\dfrac{1}{{\sqrt x }} - \dfrac{1}{{\sqrt x - 1}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x - 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{x - 4 - x + 1}}\\
= \dfrac{{ - 1}}{{\sqrt x }}.\dfrac{{\sqrt x - 2}}{{ - 3}}\\
= \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b)A = 0\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{3\sqrt x }} = 0\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {ktm} \right)
\end{array}$
Vậy ko có x thỏa mãn để $A = 0$
