ĐK : \(x\ge0\)
Ta có :
\(A=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\left(\sqrt{x}\right)^3+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+1-\sqrt{x}\)
\(=x+1\)
Lại có : \(A=2\Leftrightarrow x+1=2\Leftrightarrow x=-1\)
Vậy..