Để \(A=\dfrac{3}{x+\sqrt{x}+1}\) là số nguyên thì \(x+\sqrt{x}+1\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x+\sqrt{x}+1\in\left\{1;3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x}=0\\x+\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)