Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1\\
M = \dfrac{{x + 3\sqrt x + 5}}{{x + \sqrt x - 2}} - \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} - \dfrac{{\sqrt x - 2}}{{1 - \sqrt x }}\\
= \dfrac{{x + 3\sqrt x + 5 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 5 - x + 1 + x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b)M = \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 2}}{{\sqrt x - 1}}\\
= 1 + \dfrac{2}{{\sqrt x - 1}}\\
M \in Z\\
\dfrac{2}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x - 1} \right) \in \left\{ { - 1;1;2} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;2;3} \right\}\\
\Leftrightarrow x \in \left\{ {0;4;9} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;4;9} \right\}
\end{array}$
