. . . . . . . . .

Đáp án:

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