Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(A^2=(2x+\sqrt{4-2x^2})^2\leq [2x^2+(4-2x^2)](2+1)\)
\(\Leftrightarrow A^2\leq 12\Rightarrow -2\sqrt{3}\leq A\leq 2\sqrt{3}\)
Vậy \(A_{\max}=2\sqrt{3}\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} 2x+\sqrt{4-2x^2}=2\sqrt{3}\\ \frac{\sqrt{2x^2}}{\sqrt{2}}=\frac{\sqrt{4-2x^2}}{1}\end{matrix}\right.\Leftrightarrow x=\frac{2}{\sqrt{3}}\)