Đáp án:

$\begin{array}{l}
a)Dkxd:x > 0;x \ne 4\\
A = \left( {\dfrac{2}{{\sqrt x  - 2}} + \dfrac{3}{{2\sqrt x  + 1}} - \dfrac{{5\sqrt x  - 7}}{{2x - 3\sqrt x  - 2}}} \right)\\
:\dfrac{{2\sqrt x  + 3}}{{5x - 10\sqrt x }}\\
 = \dfrac{{2\left( {2\sqrt x  + 1} \right) + 3\left( {\sqrt x  - 2} \right) - 5\sqrt x  + 7}}{{\left( {\sqrt x  - 2} \right)\left( {2\sqrt x  + 1} \right)}}\\
.\dfrac{{5\sqrt x \left( {\sqrt x  - 2} \right)}}{{2\sqrt x  + 3}}\\
 = \dfrac{{4\sqrt x  + 2 + 3\sqrt x  - 6 - 5\sqrt x  + 7}}{{2\sqrt x  + 1}}.\dfrac{{5\sqrt x }}{{2\sqrt x  + 3}}\\
 = \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  + 1}}.\dfrac{{5\sqrt x }}{{2\sqrt x  + 3}}\\
 = \dfrac{{5\sqrt x }}{{2\sqrt x  + 1}}\\
b)A = \dfrac{{5\sqrt x }}{{2\sqrt x  + 1}} \in Z\\
 \Leftrightarrow 5\sqrt x  \vdots \left( {2\sqrt x  + 1} \right)\\
 \Leftrightarrow 2.5\sqrt x  \vdots \left( {2\sqrt x  + 1} \right)\\
 \Leftrightarrow 5\left( {2\sqrt x  + 1} \right) - 5 \vdots \left( {2\sqrt x  + 1} \right)\\
 \Leftrightarrow 5 \vdots \left( {2\sqrt x  + 1} \right)\\
 \Leftrightarrow \left( {2\sqrt x  + 1} \right) \in \left\{ {1;5} \right\}\\
 \Leftrightarrow 2\sqrt x  \in \left\{ {0;4} \right\}\\
 \Leftrightarrow \sqrt x  \in \left\{ {0;2} \right\}\\
 \Leftrightarrow x \in \left\{ {0;4} \right\}\\
Do:x \ge 0;x \ne 4\\
 \Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$