a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x-\sqrt{x}+1\ne0\\\sqrt{x}\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\\x\ne0\end{matrix}\right.\)
=> \(x>0\)
Ta có : \(A=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
=> \(A=\frac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}-1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
=> \(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
=> \(A=\sqrt{x}\left(\sqrt{x}+1\right)-1-\left(2\sqrt{x}+1\right)\)
=> \(A=x+\sqrt{x}-1-2\sqrt{x}-1\)
=> \(A=x-\sqrt{x}-2\)
b, Ta có : \(A=x-\sqrt{x}-2\)
=> \(A=x-2\sqrt{x}\frac{1}{2}+\frac{1}{4}-\frac{9}{4}\)
=> \(A=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\)
Ta thấy : \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)
=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\) với mọi x > 0 .
Vậy GTNN của A là \(-\frac{9}{4}\). Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) <=> \(x=\frac{1}{4}\)
