a)
\(A=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\\ =\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\left(2\sqrt{x}+1\right)\\ =x+\sqrt{x}+1-2\sqrt{x}-1\\ =x-\sqrt{x}\)
b)
\(A=2\Leftrightarrow x-\sqrt{x}=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow x+\sqrt{x}-2\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\\ \Leftrightarrow x=4\left(tm\right)\)
c)
\(x-\sqrt{x}=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2-\frac{1}{4}\\ =\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)
Vậy Min A = \(-\frac{1}{4}khix=\frac{1}{4}\)
