\(\text{Ta có : }x+y=1\Rightarrow\left\{{}\begin{matrix}1-y=x\\y-1=-x\end{matrix}\right.\left(1\right)\\ \)
\(A=x^2+xy-x+xy^2+y^3-y^2+xy\)
\(A=\left(x^2+xy\right)-\left(x-xy\right)+\left(y^3-y^2\right)+xy^2\)
\(A=x\left(x+y\right)-x\left(1-y\right)+y^2\left(y-1\right)+xy^2\)
Thay \(\left(1\right)\) vào suy ra :
\(A=x\left(1\right)-x\left(x\right)+y^2\left(-x\right)+xy^2\)
\(A=x-x^2+\left(-xy^2\right)+xy^2\)
\(A=x-x^2-xy^2+xy^2\)
\(A=x-x^2-\left(xy^2-xy^2\right)\)
\(A=x-x^2\)
Mà \(x^2\ge0\)
\(\Rightarrow A=x-x^2\le x\)
Dấu \("="\) xảy ra khi : \(x^2=0\Rightarrow x=0\)
\(\Rightarrow A=x-x^2\le0\)
Vậy \(A_{\left(max\right)}=0\) khi \(x=0\)
