Question

[Đàm Thảo Anh]

tìm Max A= 2x+\(\sqrt{4-2x^2}\) với -\(\sqrt{2}\) =< x=<\(\sqrt{2}\)

Answer 1

[Neet]

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

áp dụng BĐT bunhiacopxki,ta có:

\(A^2\le\left(2+1\right)\left(2x^2+4-2x^2\right)=3.4=12\)

\(\Leftrightarrow A\le\sqrt{12}\)

dấu = xảy ra khi \(\frac{\sqrt{2}}{\sqrt{2}x}=\frac{1}{\sqrt{4-2x^2}}\Leftrightarrow4-2x^2=x^2\Leftrightarrow x=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}\)

vậy Amax = \(\sqrt{12}\)khi x=\(\frac{2}{\sqrt{3}}\)