ĐKXĐ:...
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-x-3-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{-x-4}\right)\)
\(=\frac{-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(-x-4\right)}=\frac{4\sqrt{x}}{x+4}\)
Do \(x\ge0\Rightarrow\frac{4\sqrt{x}}{x+4}\ge0\)
Mặt khác \(\frac{4\sqrt{x}}{x+4}-1=\frac{-\left(x-4\sqrt{x}+4\right)}{x+4}=\frac{-\left(\sqrt{x}-2\right)^2}{x+4}\le0\) \(\forall x\ge0\)
\(\Rightarrow A\le1\Rightarrow0\le A\le1\)
Do A nguyên \(\Rightarrow\left[{}\begin{matrix}A=0\\A=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
