ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để K là số nguyên dương thì \(\left\{{}\begin{matrix}\sqrt{x}⋮\sqrt{x}-3\\\dfrac{\sqrt{x}}{\sqrt{x}-3}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-3+3⋮\sqrt{x}-3\\\sqrt{x}-3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3⋮\sqrt{x}-3\\\sqrt{x}-3>0\end{matrix}\right.\)
=>\(\sqrt{x}-3\in\left\{1;3\right\}\)
=>\(\sqrt{x}\in\left\{4;6\right\}\)
=>\(x\in\left\{14;36\right\}\)
\(K=\dfrac{\sqrt{x}-3+3}{\sqrt{x}-3}=1+\dfrac{3}{\sqrt{x}-3}\)
K nguyên khi \(\dfrac{3}{\sqrt{x}-3}\) nguyên \(\Rightarrow\sqrt{x}-3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x=\left\{0;4;16;36\right\}\)
Thay vào thấy \(x=\left\{16;36\right\}\) thỏa mãn
