ĐKXĐ: \(x\ge0\)
\(Q=\dfrac{3\sqrt{x}+15}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+3\right)+6}{\sqrt{x}+3}=3+\dfrac{6}{\sqrt{x}+3}\)
Để \(Q\in Z\Rightarrow\sqrt{x}+3\inƯ\left(6\right)\) mà \(\sqrt{x}+3\ge3\Rightarrow\sqrt{x}+3\in\left\{3;6\right\}\)
\(\Rightarrow x\in\left\{0;9\right\}\)