Ta có: \(2x+xy=4\)
\(\Leftrightarrow2x^2+x^2y=4x\)
\(\Leftrightarrow x^2y=4x-2x^2=-2\left(x^2-2x\right)\)
\(=-2\left(x^2-2x+1-1\right)\)
\(=-2\left[\left(x-1\right)^2-1\right]\)
\(=-2\left(x-1\right)^2+2\le2\)
Vậy \(A_{max}=2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
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Áp dụng BĐT AM-GM:\(4=2x+xy\ge2\sqrt{2x^2y}\Rightarrow\sqrt{x^2y}\le\frac{4}{2\sqrt{2}}\Rightarrow x^2y\le2\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}2x=xy\\2x+xy=4\end{cases}}\Leftrightarrow x=1\Rightarrow y=2\)
Vậy ...