Đáp án:

$\begin{array}{l}
a)Dk{\rm{xd}}:x \ge 0;x \ne 1\\
A = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} + \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{3\sqrt x  + 1}}{{x - 1}}\\
 = \dfrac{{{{\left( {\sqrt x  + 1} \right)}^2} + {{\left( {\sqrt x  - 1} \right)}^2} - 3\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2\sqrt x  + 1 + x - 2\sqrt x  + 1 - 3\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{2x - 3\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{2x - 2\sqrt x  - \sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 1} \right)\left( {2\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}}\\
b)x = 9\left( {tmdk} \right)\\
 \Rightarrow \sqrt x  = 3\\
 \Rightarrow A = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = \dfrac{{2.3 - 1}}{{3 + 1}} = \dfrac{5}{4}\\
c)A = 1\\
 \Rightarrow \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = 1\\
 \Rightarrow 2\sqrt x  - 1 = \sqrt x  + 1\\
 \Rightarrow \sqrt x  = 2\\
 \Rightarrow x = 4\left( {tmdk} \right)\\
d)A = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = \dfrac{{2\sqrt x  + 2 - 3}}{{\sqrt x  + 1}}\\
 = 2 - \dfrac{3}{{\sqrt x  + 1}}\\
A \in Z\\
 \Rightarrow \dfrac{3}{{\sqrt x  + 1}} \in Z\\
 \Rightarrow \left[ \begin{array}{l}
\left( {\sqrt x  + 1} \right) = 1\\
\sqrt x  + 1 = 3
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt x  = 0\\
\sqrt x  = 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 4\left( {tm} \right)
\end{array} \right.\\
e)x \ge 0;x \ne 1\\
A < 1\\
 \Rightarrow \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} < 1\\
 \Rightarrow \dfrac{{2\sqrt x  - 1 - \sqrt x  - 1}}{{\sqrt x  + 1}} < 0\\
 \Rightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} < 0\\
 \Rightarrow \sqrt x  - 2 < 0\left( {do:\sqrt x  + 1 > 0} \right)\\
 \Rightarrow \sqrt x  < 2\\
 \Rightarrow x < 4\\
Vay\,0 \le x < 4;x \ne 1\\
f)A = \dfrac{{2\sqrt x  - 1}}{{\sqrt x  + 1}} = 2 - \dfrac{3}{{\sqrt x  + 1}}\\
Do:\sqrt x  + 1 \ge 1\\
 \Rightarrow \dfrac{1}{{\sqrt x  + 1}} \le 1\\
 \Rightarrow \dfrac{3}{{\sqrt x  + 1}} \le 3\\
 \Rightarrow  - \dfrac{3}{{\sqrt x  + 1}} \ge  - 3\\
 \Rightarrow 2 - \dfrac{3}{{\sqrt x  + 1}} \ge 2 - 3\\
 \Rightarrow A \ge  - 1\\
 \Rightarrow GTNN:A =  - 1\,Khi:x = 0
\end{array}$