\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)
Vì \(x\le3\Rightarrow\dfrac{2}{\sqrt{x}}\ge\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow-\dfrac{2}{\sqrt{x}}\le-\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{3}}\le1-\dfrac{2}{\sqrt{3}}\)
\(\Rightarrow\)\(P\le\dfrac{3-2\sqrt{3}}{3}\)
Dấu = xra khi x=3
Vậy \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)