ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{1-\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
Để A nguyên thì \(2⋮x+\sqrt{x}+1\)
mà \(x+\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ
nên \(x+\sqrt{x}+1\in\left\{1;2\right\}\)
=>\(x+\sqrt{x}\in\left\{0;1\right\}\)
=>\(\left[{}\begin{matrix}x+\sqrt{x}=0\\x+\sqrt{x}-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
