a) \(A=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)
\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(A=x-\sqrt{x}-2\sqrt{x}+1+2\sqrt{x}+2\)
\(A=x-\sqrt{x}+3\)
b) Ta có:
\(A=x-\sqrt{x}+3=x-2\cdot\dfrac{1}{2}\cdot\sqrt{x}+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x-2\cdot\dfrac{1}{2}\cdot\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{11}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)
Mà: \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow A=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)
Dấu "=" xảy ra:
\(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
Vậy: \(A_{min}=\dfrac{11}{4}\Leftrightarrow x=\dfrac{1}{4}\)
