a) đkxđ : x \(\ge\) 0
A = \(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\) = \(\dfrac{x^2+\sqrt{x}+x-\sqrt{x}+1}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
= \(\dfrac{x^2+x+1}{x-\sqrt{x}+1}-\dfrac{2\sqrt{x}+1}{1}\) = \(\dfrac{x^2+x+1-\left(2\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
= \(\dfrac{x^2+x+1-\left(2x\sqrt{x}-2x+2\sqrt{x}+x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
= \(\dfrac{x^2+x+1-2x\sqrt{x}+2x-2\sqrt{x}-x+\sqrt{x}-1}{x-\sqrt{x}+1}\)
= \(\dfrac{x^2-2x\sqrt{x}+2x-\sqrt{x}}{x-\sqrt{x}+1}\) = \(\dfrac{\left(x-\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{x-\sqrt{x}+1}\) = \(x-\sqrt{x}\)
b) ta có A = 2 \(\Leftrightarrow\) \(x-\sqrt{x}=2\) \(\Leftrightarrow\) \(x-\sqrt{x}-2=0\)
giải phương trình ta có : x = 4
c) ta có đkxđ là \(x\ge0\) \(\Leftrightarrow\) \(x-\sqrt{x}\ge0\)
vậy minA = 0 khi \(x-\sqrt{x}=0\) \(\Leftrightarrow\) \(\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
