\(\dfrac{x}{3\sqrt{x}+3}\in Z\Rightarrow\dfrac{3x}{3\sqrt{x}+3}\in Z\Rightarrow\dfrac{x}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\dfrac{x-1+1}{\sqrt{x}+1}\in Z\Rightarrow\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}-1+\dfrac{1}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\sqrt{x}+1=Ư\left(1\right)=1\) (do \(\sqrt{x}+1>0\))
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
Thử lại thấy thỏa mãn, vậy \(x=0\)