Điều kiện: x\(\ge\)0
Ta có: \(\frac{3\sqrt{x}+11}{\sqrt{x}+2}=\frac{3\left(\sqrt{x}+2\right)+5}{\sqrt{x}+2}\)
\(=3+\frac{5}{\sqrt{x}+2}\)
Để \(\frac{3\sqrt{x}+11}{\sqrt{x}+2}\) là số nguyên thì \(\sqrt{x}+2\) là Ư(5)
\(\Rightarrow\sqrt{x}+2\in\){\(\pm\)1;\(\pm\)5}
\(\sqrt{x}+2=1\Leftrightarrow\sqrt{x}=-1\) (KTM)
\(\sqrt{x}+2=-1\Leftrightarrow\sqrt{x}=-3\) (KTM)
\(\sqrt{x}+2=5\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (TM)
\(\sqrt{x}+2=-5\Leftrightarrow\sqrt{x}=-7\) (KTM)
Vậy...
Đặt \(A=\frac{3\sqrt{x}+11}{\sqrt{x}+2}\Rightarrow A\sqrt{x}+2A=3\sqrt{x}+11\)
\(\Rightarrow\left(A-3\right)\sqrt{x}=11-2A\)
Với \(A=3\) ko thỏa mãn \(\Rightarrow\sqrt{x}=\frac{11-2A}{A-3}\)
Mà \(\sqrt{x}\ge0\Rightarrow\frac{11-2A}{A-3}\ge0\Rightarrow3< A\le\frac{11}{2}\)
\(A\in Z\Rightarrow A=\left\{4;5\right\}\)
\(A=4\Rightarrow\sqrt{x}=\frac{11-2A}{A-3}=3\Rightarrow x=9\)
\(A=5\Rightarrow\sqrt{x}=\frac{11-2A}{A-3}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)