Đáp án:

$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1\\
A = \left( {\dfrac{{2x + 1}}{{x\sqrt x  - 1}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\left( {1 - \dfrac{{x - 2}}{{x + \sqrt x  + 1}}} \right)\\
 = \left( {\dfrac{{2x + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\dfrac{{x + \sqrt x  + 1 - x + 2}}{{x + \sqrt x  + 1}}\\
 = \dfrac{{2x + 1 - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{{x + \sqrt x  + 1}}{{\sqrt x  + 3}}\\
 = \dfrac{{x - \sqrt x }}{{\sqrt x  - 1}}.\dfrac{1}{{\sqrt x  + 3}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 3}}\\
b)x = \dfrac{{2 - \sqrt 3 }}{2}\left( {tmdk} \right)\\
 = \dfrac{{4 - 2\sqrt 3 }}{4}\\
 = {\left( {\dfrac{{\sqrt 3  - 1}}{2}} \right)^2}\\
 \Leftrightarrow \sqrt x  = \dfrac{{\sqrt 3  - 1}}{2}\\
 \Leftrightarrow A = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} = \dfrac{{\dfrac{{\sqrt 3  - 1}}{2}}}{{\dfrac{{\sqrt 3  - 1}}{2} + 3}}\\
 = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  - 1 + 6}}\\
 = \dfrac{{\left( {\sqrt 3  - 1} \right)\left( {5 - \sqrt 3 } \right)}}{{\left( {\sqrt 3  + 5} \right)\left( {5 - \sqrt 3 } \right)}}\\
 = \dfrac{{6\sqrt 3  - 14}}{{25 - 3}}\\
 = \dfrac{{3\sqrt 3  - 7}}{{11}}\\
c)A = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} = \dfrac{{\sqrt x  + 3 - 3}}{{\sqrt x  + 3}}\\
 = 1 - \dfrac{3}{{\sqrt x  + 3}}\\
A \in Z\\
 \Leftrightarrow \dfrac{3}{{\sqrt x  + 3}} \in Z\\
 \Leftrightarrow \sqrt x  + 3 = 3\left( {do:\sqrt x  + 3 \ge 3} \right)\\
 \Leftrightarrow \sqrt x  = 0\\
 \Leftrightarrow x = 0\left( {tmdk} \right)\\
Vậy\,x = 0
\end{array}$