Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 4;x \ne 9\\
M = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)x = 11 - 6\sqrt 2 = {\left( {3 - \sqrt 2 } \right)^2}\\
\Rightarrow \sqrt x = 3 - \sqrt 2 \\
\Rightarrow M = \dfrac{{3 - \sqrt 2 + 1}}{{3 - \sqrt 2 - 3}} = \dfrac{{4 - \sqrt 2 }}{{ - \sqrt 2 }} = 1 - 2\sqrt 2 \\
c)M = 2\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = 2\\
\Rightarrow \sqrt x + 1 = 2\sqrt x - 6\\
\Rightarrow \sqrt x = 7\\
\Rightarrow x = 49\left( {tmdk} \right)\\
Vậy\,x = 49\\
d)M < 1\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} - 1 < 0\\
\Rightarrow \dfrac{{\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\Rightarrow \dfrac{4}{{\sqrt x - 3}} < 0\\
\Rightarrow \sqrt x - 3 < 0\\
\Rightarrow \sqrt x < 3\\
\Rightarrow x < 9\\
Vậy\,0 \le x < 9;x \ne 4\\
e)M = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}}\\
M \in Z\\
\Rightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in \left\{ { - 4; - 2; - 1;1;2;4} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 1;1;2;4;5;7} \right\}\\
Do:\sqrt x \ge 0;\sqrt x \ne 2;\sqrt x \ne 3\\
\Rightarrow \sqrt x \in \left\{ {1;4;5;7} \right\}\\
\Rightarrow x \in \left\{ {1;16;25;49} \right\}
\end{array}$
