Question

Tính GTNN:
A= 2+ \(\sqrt{x^2-2x+8}\)

Answer 1

Ta có : \(x^2-2x+8=\left(x^2-2x+1\right)+7=\left(x-1\right)^2+7\ge7\)

\(\Rightarrow\sqrt{x^2-2x+8}\ge\sqrt{7}\)\(\Rightarrow2+\sqrt{x^2-2x+8}\ge2+\sqrt{7}\)

\(\Rightarrow A\ge2+\sqrt{7}\). Dấu "=" xảy ra khi x = 1

Vậy Min A = \(2+\sqrt{7}\), khi x = 1

Answer 2

ta có \(\sqrt{x^2-2x+8}=\sqrt{\left(x-1\right)^2+7}\ge\sqrt{7}\)

suy ra \(2+\sqrt{x^2-2x+8}\ge2+\sqrt{7}\)

dấu = xảy ra khi và chỉ khi x-1 = 0 ---> x=1