\(A=\frac{1}{\sqrt{x}-3}\)
Dể A có giá trị là nguyên thì
\(1⋮\sqrt{x}-3\)
=> \(\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> \(\left[{}\begin{matrix}\sqrt{x}-3=1\\\sqrt{x}-3=-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x}=1+3=4\\\sqrt{x}=-1+3=2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4^2=16\\x=2^2=4\end{matrix}\right.\)
Vậy \(x\in\left\{16;4\right\}\)