Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 4\\
A = \left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{\sqrt x + 2}}} \right).{\left( {\dfrac{{\sqrt x + 2}}{2}} \right)^2}\\
= \dfrac{{\sqrt x + 2 - \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{4}\\
= \dfrac{4}{{\sqrt x - 2}}.\dfrac{{\sqrt x + 2}}{4}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}}\\
b)A = \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} = \dfrac{3}{2}\\
\Leftrightarrow 2\sqrt x + 4 = 3\sqrt x - 6\\
\Leftrightarrow \sqrt x = 10\\
\Leftrightarrow x = 100\left( {tmdk} \right)\\
Vậy\,x = 100\\
c)A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 + 4}}{{\sqrt x - 2}}\\
= 1 + \dfrac{4}{{\sqrt x - 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{4}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \left( {\sqrt x - 2} \right) \in \left\{ { - 2; - 1;1;2;4} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {0;1;3;4;6} \right\}\\
\Leftrightarrow x \in \left\{ {0;1;9;16;36} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {0;1;9;16;36} \right\}\\
d)A > 1\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - 1 > 0\\
\Leftrightarrow \dfrac{{\sqrt x + 2 - \sqrt x + 2}}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \dfrac{4}{{\sqrt x - 2}} > 0\\
\Leftrightarrow \sqrt x - 2 > 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4\\
Vậy\,x > 4\\
e)A < 1\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - 1 < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 2 - \sqrt x + 2}}{{\sqrt x - 2}} < 0\\
\Leftrightarrow \sqrt x - 2 < 0\\
\Leftrightarrow \sqrt x < 2\\
\Leftrightarrow x < 4\\
Vậy\,0 \le x < 4
\end{array}$
