Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 4\\
P = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}}\\
\Leftrightarrow \dfrac{1}{P} = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}}\\
= 1 - \dfrac{3}{{\sqrt x + 1}}\\
\dfrac{1}{P} \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;3} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Leftrightarrow x \in \left\{ {0;4} \right\}\\
Do:x \ne 4\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
2)\\
Q = \dfrac{{x - 5}}{{\sqrt x - 3}} = \dfrac{{x - 9 + 4}}{{\sqrt x - 3}}\\
= \sqrt x + 3 + \dfrac{4}{{\sqrt x - 3}}\\
Q \in Z\\
\Leftrightarrow \dfrac{4}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in \left\{ { - 2; - 1;1;2;4} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {1;2;4;5;7} \right\}\\
\Leftrightarrow x \in \left\{ {1;4;16;25;49} \right\}\\
Vậy\,x \in \left\{ {1;4;16;25;49} \right\}\\
3)Dkxd:x \ge 0;x \ne \dfrac{{25}}{4}\\
A.B\\
= \dfrac{{2\sqrt x - 5}}{{\sqrt x + 2}}.\dfrac{{2\sqrt x - 1}}{{2\sqrt x - 5}}\\
= \dfrac{{2\sqrt x - 1}}{{\sqrt x + 2}} > \dfrac{3}{4}\\
\Leftrightarrow \dfrac{{2\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{3}{4} > 0\\
\Leftrightarrow \dfrac{{8\sqrt x - 4 - 3\sqrt x - 6}}{{4\left( {\sqrt x + 2} \right)}} > 0\\
\Leftrightarrow 5\sqrt x - 10 > 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4\\
Vậy\,x > 4;x \ne \dfrac{{25}}{4}
\end{array}$
