\(D=\dfrac{x+\sqrt{x}+7}{\sqrt{x}-1}\)
\(ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(D=\dfrac{x-\sqrt{x}+2\sqrt{x}+7}{\sqrt{x}-1}\)
\(D=\dfrac{x-\sqrt{x}+2\sqrt{x}-2+9}{\sqrt{x}-1}\)
\(D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+9}{\sqrt{x}-1}\)
\(D=\sqrt{x}+2+\dfrac{9}{\sqrt{x}-1}\)
Để `D` nguyên thì \(\left\{{}\begin{matrix}\sqrt{x}\in Z\\\dfrac{9}{\sqrt{x}-1}\in Z\end{matrix}\right.\)
\(\dfrac{9}{\sqrt{x}-1}\in Z\) hay \(\sqrt{x}-1\in U\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
Mà \(x\ge0\) \(\Rightarrow\sqrt{x}-1\in\left\{1;3;9\right\}\)
`@`\(\sqrt{x}-1=1\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(tm\right)\)
`@`\(\sqrt{x}-1=3\Rightarrow\sqrt{x}=4\Rightarrow x=16\left(tm\right)\)
`@`\(\sqrt{x}-1=9\Rightarrow\sqrt{x}=10\Rightarrow x=100\left(tm\right)\)
Vậy \(x\in\left\{4;16;100\right\}\) thì `D` nguyên
