Question

Với \(x\ge-\dfrac{1}{2}\)

Tìm GTLN của \(P=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)

Answer 1

\(P=\sqrt{\left(x+2\right)\left(2x+1\right)}+2\sqrt{x+3}-2x\)

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)