Đáp án:

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)A = \left( {\frac{{2x + 1}}{{x\sqrt x  - 1}} - \frac{1}{{\sqrt x  - 1}}} \right):\left( {1 - \frac{{x - 2}}{{x + \sqrt x  + 1}}} \right)\\
 = \frac{{2x + 1 - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\left( {\frac{{x + \sqrt x  + 1 - x + 2}}{{x + \sqrt x  + 1}}} \right)\\
 = \frac{{2x + 1 - x - \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\frac{{x + \sqrt x  + 1}}{{\sqrt x  + 3}}\\
 = \frac{{x - \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \frac{{\sqrt x }}{{\sqrt x  + 3}}\\
b)x = \frac{{2 - \sqrt 3 }}{2}\left( {tmdk} \right)\\
x = \frac{{4 - 2\sqrt 3 }}{4} = \frac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{{{2^2}}} = {\left( {\frac{{\sqrt 3  - 1}}{2}} \right)^2}\\
 \Rightarrow \sqrt x  = \frac{{\sqrt 3  - 1}}{2} = \frac{{3 - 1}}{{2\left( {\sqrt 3  + 1} \right)}} = \frac{1}{{\sqrt 3  + 1}}\\
 \Rightarrow A = \frac{{\sqrt x }}{{\sqrt x  + 3}} = \frac{{\frac{1}{{\sqrt 3  + 1}}}}{{\frac{1}{{\sqrt 3  + 1}} + 3}} = \frac{1}{{1 + 3\sqrt 3  + 3}} = \frac{{3\sqrt 3  - 4}}{{{{\left( {3\sqrt 3 } \right)}^2} - {4^2}}} = \frac{{3\sqrt 3  - 4}}{9}\\
c)A = \frac{{\sqrt x }}{{\sqrt x  + 3}} = 1 - \frac{3}{{\sqrt x  + 3}}\\
A \in Z\\
 \Rightarrow \frac{3}{{\sqrt x  + 3}} \in Z\\
 \Rightarrow \left( {\sqrt x  + 3} \right) \in U\left( 3 \right)\\
Do:\sqrt x  + 3 \ge 3\forall x \ge 0\\
 \Rightarrow \sqrt x  + 3 = 3\\
 \Rightarrow x = 0\left( {tmdk} \right)\\
d)A = 1 - \frac{3}{{\sqrt x  + 3}}\\
Do:\sqrt x  + 3 \ge 3\\
 \Rightarrow \frac{3}{{\sqrt x  + 3}} \le \frac{3}{3} = 1\\
 \Rightarrow  - \frac{3}{{\sqrt x  + 3}} \ge  - 1\\
 \Rightarrow 1 - \frac{3}{{\sqrt x  + 3}} \ge 0\\
 \Rightarrow GTNN:A = 0 \Leftrightarrow x = 0\\
e)A = \frac{{\sqrt x }}{{\sqrt x  + 3}} < 1\left( {do:\sqrt x  \ge 0} \right)\\
f)A > \frac{1}{2}\\
 \Rightarrow \frac{{\sqrt x }}{{\sqrt x  + 3}} > \frac{1}{2}\\
 \Rightarrow 2\sqrt x  > \sqrt x  + 3\\
 \Rightarrow \sqrt x  > 3\\
 \Rightarrow x > 9\\
g)A = \frac{1}{3}\\
 \Rightarrow \frac{{\sqrt x }}{{\sqrt x  + 3}} = \frac{1}{3}\\
 \Rightarrow 3\sqrt x  = \sqrt x  + 3\\
 \Rightarrow 2\sqrt x  = 3\\
 \Rightarrow \sqrt x  = \frac{3}{2}\\
 \Rightarrow x = \frac{9}{4}\left( {tmdk} \right)
\end{array}$