Question

Tìm max \(y=\dfrac{x}{x^2+2}\) biết \(x>0\)

Answer 1

ĐK: y\(\ge0\)

\(y=\dfrac{x}{x^2+2}\)

\(\Rightarrow yx^2-x+2y=0\)

Để pt có nghiệm thì \(\Delta\ge0\)

\(\Rightarrow\left(-1\right)^2-4y.2y\ge0\)

\(\Leftrightarrow1-8y^2\ge0\)

\(\Leftrightarrow y^2\le\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{-\sqrt{2}}{4}\le y\le\dfrac{\sqrt{2}}{4}\)

Vậy y_max=\(\dfrac{\sqrt{2}}{4}\Leftrightarrow\dfrac{x}{x^2+2}=\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow\sqrt{2}x^2+2\sqrt{2}-4x=0\)

\(\Leftrightarrow x=\sqrt{2}\left(TM\right)\)