Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 4;x \ne 9\\
Q = \left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{5\sqrt x + 2}}{{4 - x}}} \right):\dfrac{{3\sqrt x - x}}{{x + 4\sqrt x + 4}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) - 2\sqrt x \left( {\sqrt x - 2} \right) - 5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\\
\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 - 2x + 4\sqrt x - 5\sqrt x - 2}}{{\sqrt x - 2}}.\dfrac{{\sqrt x + 2}}{{\sqrt x .\left( {3 - \sqrt x } \right)}}\\
= \dfrac{{ - x + 2\sqrt x }}{{\sqrt x - 2}}.\dfrac{{\sqrt x + 2}}{{\sqrt x \left( {3 - \sqrt x } \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x - 3 + 5}}{{\sqrt x - 3}} = 1 + \dfrac{5}{{\sqrt x - 3}}\\
Q \in Z\\
\Rightarrow \dfrac{5}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in \left\{ { - 1;1;5} \right\}\left( {do:\sqrt x - 3 \ge - 3} \right)\\
\Rightarrow \sqrt x \in \left\{ {2;4;8} \right\}\\
\Rightarrow x \in \left\{ {4;16;64} \right\}\\
Do:x \ne 4\\
\Rightarrow x \in \left\{ {16;64} \right\}
\end{array}$
Vậy x=16 hoặc x=64
