Question

tìm min

A =    \(x-2\sqrt{x-4}+3\)

B = \(\sqrt{3x^2-12x+16}+\sqrt{x^4-8x^2+17}\)

Answer 1

\(A=x-4-2\sqrt{x-4}+1+6=\left(\sqrt{x-4}-1\right)^2+6\ge6\)

dấu \(=\)xảy ra khi \(\sqrt{x-4}=1\Leftrightarrow x=5\)

Answer 2

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

Dấu \(=\)xảy ra khi \(x=2\)