Question

Cho \(x\ge-\frac{1}{2}\) Tìm GTLN của:

\(A=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)

Answer 1

\(A=\sqrt{2x^2+5x+2}+2\sqrt{x+3}-2x\)

\(2A=2\sqrt{2x^2+5x+2}+4\sqrt{x+3}-4x\)

\(2A=2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)

\(\le2x+1+x+2+4+x+3-4x=10\)

=>2A\(\le10\Rightarrow A\le5\)

dấu bằng xảy ra \(\Leftrightarrow2x+1=x+2\)

và x+3=4

=>x=1

maxA=5 khi x=1

 

 

Answer 2

Khó vậy ta ????