Đáp án:

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)A = \left( {\dfrac{{\sqrt x }}{{x\sqrt x  - 1}} + \dfrac{1}{{\sqrt x  - 1}}} \right):\dfrac{{\sqrt x  + 1}}{{x + \sqrt x  + 1}}\\
 = \dfrac{{\sqrt x  + x + \sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{x + \sqrt x  + 1}}{{\sqrt x  + 1}}\\
 = \dfrac{{x + 2\sqrt x  + 1}}{{\sqrt x  - 1}}.\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
b)x = \sqrt {3 + 2\sqrt 2 }  - \sqrt {3 - 2\sqrt 2 } \\
 = \sqrt {{{\left( {\sqrt 2  + 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt 2  - 1} \right)}^2}} \\
 = \sqrt 2  + 1 - \left( {\sqrt 2  - 1} \right)\\
 = \sqrt 2  + 1 - \sqrt 2  + 1\\
 = 2\\
 \Rightarrow \sqrt x  = \sqrt 2 \\
 \Rightarrow A = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt 2  + 1}}{{\sqrt 2  - 1}}\\
 = \dfrac{{{{\left( {\sqrt 2  + 1} \right)}^2}}}{{2 - 1}}\\
 = 3 + 2\sqrt 2 \\
c)A > 1\\
 \Rightarrow \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} > 1\\
 \Rightarrow \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} - 1 > 0\\
 \Rightarrow \dfrac{{\sqrt x  + 1 - \sqrt x  + 1}}{{\sqrt x  - 1}} > 0\\
 \Rightarrow \dfrac{2}{{\sqrt x  - 1}} > 0\\
 \Rightarrow \sqrt x  - 1 > 0\\
 \Rightarrow \sqrt x  > 1\\
 \Rightarrow x > 1\\
\text{Vậy}\,x > 1\\
d)x > 1 \Rightarrow A > 1\\
 \Rightarrow A > \sqrt A \\
e)A = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  - 1 + 2}}{{\sqrt x  - 1}} = 1 + \dfrac{2}{{\sqrt x  - 1}}\\
A \in Z\\
 \Rightarrow 2 \vdots \left( {\sqrt x  - 1} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt x  - 1 =  - 1\\
\sqrt x  - 1 = 1\\
\sqrt x  - 1 = 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt x  = 0\\
\sqrt x  = 2\\
\sqrt x  = 3
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 4\left( {tm} \right)\\
x = 9\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x \in \left\{ {0;4;9} \right\}
\end{array}$