Question

Tìm max của: \(\dfrac{\sqrt{x}}{x-2\sqrt{x}+9}\)

Answer 1

\(\forall x\in R\Rightarrow A=\dfrac{\sqrt{x}}{x-2\sqrt{x}+9}\Leftrightarrow A\left(x-2\sqrt{x}+9\right)=\sqrt{x}\)

\(\Leftrightarrow Ax-2A\sqrt{x}-\sqrt{x}+9A=0\)

\(\Leftrightarrow A\sqrt{x}^2-\sqrt{x}\left(2A+1\right)+9A=0\)

\(\Rightarrow\Delta\ge0\Rightarrow\left(2A+1\right)^2-36A^2=-32A^2+4A+1\ge0\Rightarrow-\dfrac{1}{8}\le A\le\dfrac{1}{4}\Rightarrow A\le\dfrac{1}{4}\Rightarrow MaxA=\dfrac{1}{4}\)

\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=9\)