Đáp án:
$\begin{array}{l}
x \ge 0;x \ne 4\\
a)A = \left( {\frac{2}{{2 - \sqrt x }} + \frac{1}{{\sqrt x + 2}} - \frac{{\sqrt x }}{{x - 4}}} \right):\left( {\frac{{6 - x}}{{\sqrt x + 2}} + \sqrt x - 2} \right)\\
= \left( {\frac{{ - 2}}{{\sqrt x - 2}} + \frac{1}{{\sqrt x + 2}} - \frac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right):\left( {\frac{{6 - x + \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}} \right)\\
= \frac{{ - 2\left( {\sqrt x + 2} \right) + \sqrt x - 2 - \sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x + 2}}{{6 - x + x - 4}}\\
= \frac{{ - 2\sqrt x - 6}}{{\sqrt x - 2}}.\frac{1}{2}\\
= \frac{{ - \sqrt x - 3}}{{\sqrt x - 2}}\\
b)A = \frac{{ - \sqrt x - 3}}{{\sqrt x - 2}} = \frac{{ - \left( {\sqrt x - 2} \right) - 5}}{{\sqrt x - 2}} = - 1 - \frac{5}{{\sqrt x - 2}}\\
A \in Z \Rightarrow \frac{5}{{\sqrt x - 2}} \in Z\\
\Rightarrow \left( {\sqrt x - 2} \right) \in U\left( 5 \right) = {\rm{\{ }} - 5; - 1;1;5\} \\
\Rightarrow \sqrt x \in {\rm{\{ }} - 3;1;3;7\} \left( {loại\, - 3} \right)\\
\Rightarrow x \in {\rm{\{ }}1;9;49\} \left( {tmdk} \right)
\end{array}$
