Đáp án:
$\begin{array}{l}
B1)Dkxd:x \ge 0;x \ne 4\\
1)A = 2\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 2}} = 2\\
\Leftrightarrow \sqrt x = 2\sqrt x - 4\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16\\
b)A > 2\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x - 2}} - 2 > 0\\
\Leftrightarrow \dfrac{{\sqrt x - 2\sqrt x + 4}}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \dfrac{{4 - \sqrt x }}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow 2 < \sqrt x < 4\\
\Leftrightarrow 4 < x < 16\\
Vậy\,4 < x < 16\\
c)A < 2\\
\Leftrightarrow \left[ \begin{array}{l}
0 \le x < 4\\
x > 16
\end{array} \right.\\
Vậy\,0 \le x < 4;x > 16\\
2)A = \dfrac{{\sqrt x }}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 + 2}}{{\sqrt x - 2}}\\
= 1 + \dfrac{2}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \left( {\sqrt x - 2} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;1;3;4} \right\}\\
\Leftrightarrow x \in \left\{ {0;1;9;16} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;1;9;16} \right\}\\
B2)x \ge 0\\
a)A > 3\\
\Leftrightarrow \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}} - 3 > 0\\
\Leftrightarrow \dfrac{{\sqrt x + 5 - 3\sqrt x - 6}}{{\sqrt x + 2}} > 0\\
\Leftrightarrow - 2\sqrt x - 1 > 0\\
\Leftrightarrow 2\sqrt x < - 1\\
\Leftrightarrow \sqrt x < - \dfrac{1}{2}\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
b)A < 5\\
\Leftrightarrow \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}} - 5 < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 5 - 5\sqrt x - 10}}{{\sqrt x + 2}} < 0\\
\Leftrightarrow - 4\sqrt x - 5 < 0\\
\Leftrightarrow 4\sqrt x > - 5\\
\Leftrightarrow \sqrt x > - \dfrac{5}{4}\left( {tmdk} \right)\\
Vậy\,x \ge 0
\end{array}$
