Đáp án:
$\begin{array}{l}
a)x \ge 0;x \ne 4;x \ne 9\\
Q = \dfrac{{2\sqrt x - 9}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)Q = 2\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = 2\\
\Rightarrow \sqrt x + 1 = 2\sqrt x - 6\\
\Rightarrow \sqrt x = 7\\
\Rightarrow x = 49\left( {tmdk} \right)\\
\text{Vậy}\,x = 49\\
c)Q = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}}\\
= 1 + \dfrac{4}{{\sqrt x - 3}}\\
Q \in Z\\
\Rightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in U\left( 4 \right) = \left\{ { - 4; - 2; - 1;1;2;4} \right\}\\
\Rightarrow \sqrt x \in \left\{ { - 1;1;2;4;5;7} \right\}\\
Do:\sqrt x \ge 0;\sqrt x \ne 2;\sqrt x \ne 3\\
\Rightarrow \sqrt x \in \left\{ {1;4;5;7} \right\}\\
\Rightarrow x \in \left\{ {1;16;25;49} \right\}\\
\text{Vậy}\,x \in \left\{ {1;16;25;49} \right\}
\end{array}$
