\(\dfrac{3\sqrt{x}+11}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+5}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{5}{\sqrt{x}+2}=3+\dfrac{5}{\sqrt{x}+2}\)
Để bt nguyên thì:
\(\dfrac{5}{\sqrt{x}+2}\in Z\Leftrightarrow\sqrt{x}+2\inƯ\left(5\right)\)
\(\Leftrightarrow\sqrt{x}+2=\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow\sqrt{x}=\left\{3\right\}\Leftrightarrow x=9\)
vậy........