Question

Rút gọn và tìm GTLN của

\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{3}\left(x\ge0;x\ne1\right)\)

Answer 1

\(A=\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{3}=\dfrac{-x-\sqrt{x}-1+x+2+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{3}{\sqrt{x}-1}=\dfrac{3}{x+\sqrt{x}+1}\text{≤}\dfrac{3}{1}=3\) ( x ≥ 0 ; x # 1 )

⇒ \(A_{Max}=3."="\) ⇔ \(x=0\left(TM\right)\)