Đặt \(A=\dfrac{x-2}{\sqrt{x}-3}\)
ĐKXĐ: x>=0; x<>9
Để A là số tự nhiên thì \(\left\{{}\begin{matrix}x-2⋮\sqrt{x}-3\\A>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-9+7⋮\sqrt{x}-3\\\dfrac{x-2}{\sqrt{x}-3}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7⋮\sqrt{x}-3\\\dfrac{x-2}{\sqrt{x}-3}>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-3\in\left\{1;-1;7;-7\right\}\\\dfrac{x-2}{\sqrt{x}-3}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}\in\left\{4;2;10;-4\right\}\\\dfrac{x-2}{\sqrt{x}-3}>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{16;4;100\right\}\\\dfrac{x-2}{\sqrt{x}-3}>=0\end{matrix}\right.\)
=>\(x\in\left\{16;4;100\right\}\)
