\(A=x^2+y^2+xy+3x+3y+2018\)
\(\Leftrightarrow2A=2x^2+2y^2+2xy+6x+6y+4036\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+\left(y^2+6y+9\right)+4018\)
\(=\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2+4018\)
\(\Rightarrow A=\dfrac{\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2}{2}+2009\)
Ta có : \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(x+3\right)^2\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2}{2}+2009\ge2009\)
Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x+3\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-3\)
Vậy \(Min_A=2009\Leftrightarrow x=y=-3\)
