ĐKXĐ: \(x\ge0;x\ne1\)
\(D=\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x\left(\sqrt{x}-1\right)+\sqrt{x}-1}\right):\left(\frac{x+1-\sqrt{x}}{x+1}\right)\)
\(=\left(\frac{1}{\sqrt{x}-1}-\frac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x-\sqrt{x}+1}{x+1}\right)\)
\(=\left(\frac{x+1-2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x-\sqrt{x}+1}{x+1}\right)\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(x+1\right)}{\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(\Rightarrow D=\frac{\sqrt{x}+1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\)
Do \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow D>0\) \(\forall x\in TXĐ\)
Đặt \(\sqrt{x}=a\ge0;a\ne1\) \(\Rightarrow D=\frac{a+1}{a^2-a+1}\)
\(\Leftrightarrow Da^2-Da+D=a+1\)
\(\Leftrightarrow Da^2-\left(D+1\right)a+D-1=0\)
\(\Delta=\left(D+1\right)^2-4D\left(D-1\right)=-3D^2+6D+1\ge0\)
\(\Rightarrow\frac{3-2\sqrt{3}}{3}\le D\le\frac{3+2\sqrt{3}}{3}\)
Mà D nguyên và \(D>0\) \(\Rightarrow D=\left\{1;2\right\}\)
\(D=1\Rightarrow a^2-2a=0\Rightarrow\left[{}\begin{matrix}a=0\\a=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(D=2\Rightarrow2a^2-3a+1=0\Rightarrow\left[{}\begin{matrix}a=1\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{4}\end{matrix}\right.\)
