Question

Tìm GTLN 

\(P=\dfrac{\sqrt{x}}{x+3}\)

Answer 1

ĐK: \(x\ge0\)

Ta có:

 \(x+3\ge2\sqrt{3x}\left(bđt.cosin\right)\\ \Rightarrow P=\dfrac{\sqrt{x}}{x+3}\le\dfrac{\sqrt{x}}{2\sqrt{3x}}\)

mà \(\dfrac{\sqrt{x}}{2\sqrt{3x}}=\dfrac{1}{2\sqrt{3}}\)

\(\Rightarrow GTLN\left(P\right)=\dfrac{1}{2\sqrt{3}}\) dấu = sảy ra khi x=3

Answer 2

ĐK:\(x\ge0\)

Ta có:

\(x+3\ge2\sqrt{3x}\left(bđt.cosin\right)\)

\(\Rightarrow P=\dfrac{\sqrt{x}}{x+3}\le\dfrac{\sqrt{x}}{2\sqrt{3x}}\)

Mà \(\dfrac{\sqrt{x}}{2\sqrt{3x}}=\dfrac{1}{2\sqrt{3}}\) 

\(\Rightarrow GTLN\left(P\right)=\dfrac{1}{2\sqrt{3}}\) dấu \(=\) xảy ra khi \(x=3\)