Đáp án:
$\begin{array}{l}
a)Dkxd:a > 0;a \ne 1\\
A = \left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{1}{{\sqrt a - 1}}} \right).\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{\sqrt a - 1 + \sqrt a + 1}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{2\sqrt a }}{{\sqrt a - 1}}.\dfrac{1}{{\sqrt a }}\\
= \dfrac{2}{{\sqrt a - 1}}\\
b)A < 0\\
\Rightarrow \dfrac{2}{{\sqrt a - 1}} < 0\\
\Rightarrow \sqrt a - 1 < 0\\
\Rightarrow \sqrt a < 1\\
\Rightarrow a < 1\\
\text{Vậy}\,0 < a < 1\,\text{thì}:A < 0\\
c)a = 3 + 2\sqrt 2 \\
= 2 + 2.\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\Rightarrow \sqrt a = \sqrt 2 + 1\\
\Rightarrow A = \dfrac{2}{{\sqrt a - 1}} = \dfrac{2}{{\sqrt 2 + 1 - 1}} = \dfrac{2}{{\sqrt 2 }} = \sqrt 2 \\
\text{Vậy}\,A = \sqrt 2 \\
d)A = \dfrac{2}{{\sqrt a - 1}}\\
A \in Z\\
\Rightarrow \left( {\sqrt a - 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow \sqrt a \in \left\{ { - 1;0;2;3} \right\}\\
Do:\sqrt a > 0\left( {khi:a > 0} \right)\\
\Rightarrow \sqrt a \in \left\{ {2;3} \right\}\\
\Rightarrow a \in \left\{ {4;9} \right\}\left( {tmdk} \right)\\
\text{Vậy}\,a = 4/a = 9
\end{array}$
