Đáp án:
$\begin{array}{l}
1)a)M = \left( {\frac{{a\sqrt a - 1}}{{a - \sqrt a }} - \frac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right).\frac{{a + 2}}{{a - 2}}\\
= \left[ {\frac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \frac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right].\frac{{a + 2}}{{a - 2}}\\
= \frac{{a + \sqrt a + 1 - \left( {a - \sqrt a + 1} \right)}}{{\sqrt a }}.\frac{{a + 2}}{{a - 2}}\\
= 2.\frac{{a + 2}}{{a - 2}}\\
b)a > 0;a \ne 1;a \ne 2\\
M = \frac{{2a + 4}}{{a - 2}} = \frac{{2a - 4 + 8}}{{a - 2}} = 2 + \frac{8}{{a - 2}}\\
\Rightarrow 8 \vdots \left( {a - 2} \right)\\
\Rightarrow a - 2 \in {\rm{\{ }} - 8; - 4; - 2; - 1;1;2;4;8\} \\
\Rightarrow a \in {\rm{\{ }} - 6; - 2;0;1;3;4;6;10\} \\
\Rightarrow a \in {\rm{\{ }}3;4;6;10\} \\
2)a)Q = \frac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \frac{{\sqrt x + 3}}{{\sqrt x - 2}} - \frac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \frac{{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)}}\\
= \frac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \frac{{\sqrt x + 2}}{{\sqrt x - 3}}\\
b)x \ge 0;x \ne 4;x \ne 9\\
Q < 0\\
\Rightarrow \frac{{\sqrt x + 2}}{{\sqrt x - 3}} < 0\\
\Rightarrow \sqrt x - 3 < 0\\
\Rightarrow x < 9\\
Vay\,0 \le x < 9;x \ne 4\\
c)Q < 1\\
\Rightarrow \frac{{\sqrt x + 2}}{{\sqrt x - 3}} < 1\\
\Rightarrow \frac{{\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\Rightarrow \frac{1}{{\sqrt x - 3}} < 0\\
\Rightarrow \sqrt x - 3 < 0\\
\Rightarrow x < 9\\
Vay\,0 \le x < 9;x \ne 4\\
d)x = \frac{2}{{2 + \sqrt 3 }} = \frac{{2\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}} = 4 - 2\sqrt 3 \\
\Rightarrow x = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 3 - 1\\
\Rightarrow Q = \frac{{\sqrt 3 - 1 + 2}}{{\sqrt 3 - 1 - 3}} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 4}}
\end{array}$
