Đáp án:
a. \(\dfrac{5}{{\sqrt x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \left( {\dfrac{2}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{x + 9}}{{x - 9}}} \right):\left( {\dfrac{{\sqrt x - 2}}{{\sqrt x + 3}}} \right)\\
= \left[ {\dfrac{{2\sqrt x - 6 + x + 3\sqrt x - x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right].\dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{{5\sqrt x - 15}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{5}{{\sqrt x - 2}}\\
b.A \in Z\\
\Leftrightarrow \dfrac{5}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \sqrt x - 2 \in U\left( 5 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 5\\
\sqrt x - 2 = - 5\left( l \right)\\
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 49\\
x = 9\left( l \right)\\
x = 1
\end{array} \right.
\end{array}\)
