Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 9\\
a)x = 36\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 6\\
\Leftrightarrow A = \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = \dfrac{{7.6 - 2}}{{2.6 + 1}} = \dfrac{{40}}{{13}}\\
b)A = B\\
\Leftrightarrow \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = \dfrac{{\sqrt x + 3}}{{\sqrt x - 3}} + \dfrac{{\sqrt x - 3}}{{\sqrt x + 3}} - \dfrac{{36}}{{x - 9}}\\
\Leftrightarrow \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = \dfrac{{{{\left( {\sqrt x + 3} \right)}^2} + {{\left( {\sqrt x - 3} \right)}^2} - 36}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
\Leftrightarrow \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = \dfrac{{x + 6\sqrt x + 9 + x - 6\sqrt x + 9 - 36}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
\Leftrightarrow \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = \dfrac{{2x - 18}}{{x - 9}}\\
\Leftrightarrow \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} = 2\\
\Leftrightarrow 7\sqrt x - 2 = 4\sqrt x + 2\\
\Leftrightarrow 3\sqrt x = 4\\
\Leftrightarrow \sqrt x = \dfrac{4}{3}\\
\Leftrightarrow x = \dfrac{{16}}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{16}}{9}\\
c)A = \dfrac{{7\sqrt x - 2}}{{2\sqrt x + 1}} \in N\\
\Leftrightarrow \left( {7\sqrt x - 2} \right) \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow 2.\left( {7\sqrt x - 2} \right) \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow 14\sqrt x - 4 \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow 14\sqrt x + 7 - 11 \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow 7.\left( {2\sqrt x + 1} \right) - 11 \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow 11 \vdots \left( {2\sqrt x + 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2\sqrt x + 1 = 1\\
2\sqrt x + 1 = 11
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 25
\end{array} \right.\\
+ Khi:x = 0 \Leftrightarrow A = \dfrac{{ - 2}}{1} = - 2\left( {ktm} \right)\\
+ Khi:x = 25 \Leftrightarrow A = \dfrac{{7.5 - 2}}{{2.5 + 1}} = \dfrac{{33}}{{11}} = 3\left( {tm} \right)\\
Vậy\,x = 25
\end{array}$
