\(A=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)
- Với \(x=0\Rightarrow A=0\)
- Với \(x>0\Rightarrow A>0\)
Hơn nữa ta có \(A=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}\le\frac{2}{2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}-1}=2\)
\(\Rightarrow0< A\le2\) mà A nguyên \(\Rightarrow\left[{}\begin{matrix}A=1\\A=2\end{matrix}\right.\)
- Với \(A=1\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Leftrightarrow2\sqrt{x}=x-\sqrt{x}+1\)
\(\Leftrightarrow x-3\sqrt{x}+1=0\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)
- Với\(A=2\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=2\Leftrightarrow\sqrt{x}=x-\sqrt{x}+1\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Rightarrow x=1\)
