Question

Tìm  GTLN  của  biểu  thức B=\(\sqrt{-X^2+2X+2}\)

Answer 1

\(B=\sqrt{-x^2+2x+2}=\sqrt{-x^2+2x-1+3}\)

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

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

Xảy ra khi x=1

Answer 2

Ta có \(-x^2+2x+2\)=\(-\left(x^2-2x+1-3\right)\)=\(3-\left(x-1\right)^2\le3\)

       Dấu '=' xảy ra \(\Leftrightarrow x=1\)

Do đó MaxB=\(\sqrt{3}\)(Dấu '=' xảy ra \(\Leftrightarrow x=1\))