Đáp án:
$\begin{array}{l}
A = \frac{{x - 15}}{{\sqrt x + 3}} = \frac{{x + 3\sqrt x - 3\sqrt x - 9 - 6}}{{\sqrt x + 3}}\\
= \frac{{\sqrt x \left( {\sqrt x + 3} \right) - 3\left( {\sqrt x + 3} \right) - 6}}{{\sqrt x + 3}}\\
= \sqrt x - 3 - \frac{6}{{\sqrt x + 3}}\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt x \in Z\\
\frac{6}{{\sqrt x + 3}} \in Z
\end{array} \right.
\end{array}$
$\begin{array}{l}
Do:\sqrt x + 3 \ge 3\\
\Rightarrow \sqrt x + 3 \in Ư\left( 6 \right) = \left\{ {3;6} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;3} \right\}\\
\Rightarrow x \in \left\{ {0;9} \right\}
\end{array}$
Vậy x=0 hoặc x=9
