với \(x\ge\dfrac{1}{2}\)(điều kiện chắc vậy)
A=\(x+\dfrac{4x+2}{2x-1}=\dfrac{x\left(2x-1\right)}{2x-1}+\dfrac{4x+2}{2x-1}\)
\(=\dfrac{2x^2-x+4x+2}{2x-1}=\dfrac{2x^2+3x+2}{2x-1}\)
\(=>2A=\)\(\dfrac{4x^2+6x+4}{2x-1}\)
\(=\dfrac{4x^2-4x+1+10x+3}{2x-1}\)
\(=\dfrac{\left(2x-1\right)^2+5\left(2x-1\right)+8}{2x-1}=2x-1+\dfrac{8}{2x-1}+5\)
\(\ge2\sqrt{8}+5\)
=>\(A\ge\dfrac{2\sqrt{8}+5}{2}=\sqrt{8}+\dfrac{5}{2}\)
Dấu"=" xảy ra<=>\(x=\dfrac{1}{2}\left(1+2\sqrt{2}\right)\)(TM)
Vậy min A=\(\sqrt{8}+\dfrac{5}{2}\)
Đúng 1
Bình luận (0)