Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9\\
x \ne \dfrac{9}{4}
\end{array} \right.\\
A = \left( {\dfrac{1}{{\sqrt x - 3}} + \dfrac{3}{{\sqrt x + 3}}} \right).\dfrac{{\sqrt x + 3}}{{2\sqrt x - 3}}\\
= \dfrac{{\left( {\sqrt x + 3} \right) + 3.\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{2\sqrt x - 3}}\\
= \dfrac{{4\sqrt x - 6}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{2\sqrt x - 3}}\\
= \dfrac{{2.\left( {2\sqrt x - 3} \right).\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right).\left( {2\sqrt x - 3} \right)}}\\
= \dfrac{2}{{\sqrt x - 3}}\\
b,\\
A = \dfrac{1}{2} \Leftrightarrow \dfrac{2}{{\sqrt x - 3}} = \dfrac{1}{2}\\
\Leftrightarrow \sqrt x - 3 = 4\\
\Leftrightarrow \sqrt x = 7\\
\Leftrightarrow x = 49\\
c,\\
A \in Z \Rightarrow \dfrac{2}{{\sqrt x - 3}} \in Z \Rightarrow \left( {\sqrt x - 3} \right) \in \left\{ { \pm 1; \pm 2} \right\}\\
\Rightarrow \sqrt x \in \left\{ {1;2;4;5} \right\}\\
\Rightarrow x \in \left\{ {1;4;16;25} \right\}
\end{array}\)
