Ta có :
\(2x+xy=4\)
\(\Leftrightarrow x\left(y+2\right)=4\)
\(\Leftrightarrow y+2=\frac{4}{x}\)
\(\Leftrightarrow y=\frac{4}{x}-2\) \(\left(1\right)\)
Thay (1) vào A ta có :
\(A=x^2y=x^2\left(\frac{4}{x}-2\right)=4x-2x^2\)
\(\Leftrightarrow A=2-2x^2+4x-2\)
\(\Leftrightarrow A=2-2\left(x^2-2x+1\right)\)
\(\Leftrightarrow A=2-2\left(x-1\right)^2\)
Với mọi x ta có :
\(2\left(x-1\right)^2\ge0\)
\(\Leftrightarrow-2\left(x-1\right)^2\le0\)
\(\Leftrightarrow2-2\left(x-1\right)^2\le2\)
\(\Leftrightarrow A\le2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy..