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