Cho \(x\ge2\). Tìm GTLN của \(f\left(x\right)=\dfrac{\sqrt{x-2}}{x}\).
\(\dfrac{1}{2\sqrt{2}}\).\(\sqrt{2}\).\(\dfrac{\sqrt{2}}{2}\).\(-\sqrt{2}\).Hướng dẫn giải:Có \(\dfrac{x}{\sqrt{x-2}}=\dfrac{\left(x-2\right)+2}{\sqrt{x-2}}\ge\dfrac{2\sqrt{\left(x-2\right)2}}{\sqrt{x-2}}=2\sqrt{2}\) nên \(f\left(x\right)\le\dfrac{1}{2\sqrt{2}}\)