Question

Tìm Max:

a) y=\(\sqrt{-x^2+2x+2}\)

b) y= 2-\(\sqrt{4x^2-4x+1}\)

Answer 1

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

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

Dễ thấy: \(-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2+2\le2\)

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

Đẳng thức xảy ra khi \(x=1\)

b)\(y=2-\sqrt{4x^2-4x+1}\)

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

Dễ thấy: \(\sqrt{\left(2x-1\right)^2}\ge0\Rightarrow-\sqrt{\left(2x-1\right)^2}\le0\)

\(y=2-\sqrt{4x^2-4x+1}\le2\)

Đẳng thức xảy ra khi \(x=\frac{1}{2}\)