Question

I : Tìm max

\(A=\sqrt{4-x^2}\)

B=\(\sqrt{-x^2+x+\frac{1}{4}}\)

help me !!!

Answer 1

Câu 1:

Áp dụng BĐT Cô-si:

\(A=\sqrt{\left(2-x\right)\left(2+x\right)}\le\frac{2-x+2+x}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow2-x=2+x\Leftrightarrow x=0\)

Câu 2:

\(B=\sqrt{-x^2+x+\frac{1}{4}}\)

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

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

\(B=\sqrt{-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\right]}\)

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

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)