Question

\(\sqrt{4x^2+12x+25}+\sqrt{16x^2+48x+54}\)

tìm min của b thức

Answer 1

\(\sqrt{4x^2+12x+25}+\sqrt{16x^2+48x+54}\)

\(=\sqrt{\left(2x+3\right)^2+16}+\sqrt{\left(4x+6\right)^2+18}\ge\sqrt{16}+\sqrt{18}=4+3\sqrt{2}\)

Vậy Min của BT là \(4+3\sqrt{2}\)\(\Leftrightarrow x=\frac{-3}{2}\)

Answer 2

\(\sqrt{\left(2x\right)^2+2.2x.3+9+16}+\sqrt{4\left[\left(2x\right)^2+2.2x.3+9\right]+18}...\)

\(=\sqrt{\left(2x+3\right)^2+16}+\sqrt{4\left(2x+3\right)^2+18}\)

\(\ge\sqrt{16}+\sqrt{18}=4+3\sqrt{2}.\)(do \(\left(2x+3\right)^2\ge0\))

Dấu '=' xảy ra khi \(\left(2x+3\right)^2=0\Leftrightarrow x=-\frac{3}{2}.\)