\(x^2+2xy+6x+6y+2y^2+8=0\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+9=1-y^2\)
\(\Leftrightarrow\left(x+y+3\right)^2=1-y^2\)
Ta thấy : \(1-y^2\le1\forall y\) \(\Rightarrow\left(x+y+3\right)^2\le1\)
\(\Rightarrow-1\le x+y+3\le1\)
\(\Rightarrow-1+2013\le x+y+3+2013\le1+2013\)
\(\Rightarrow2012\le x+y+2016\le2014\)
Vậy ta có :
+) Min \(B=2012\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=-1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-4\end{cases}}\)
+) Max \(M=2014\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}y=0\\x+y+3=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=-2\end{cases}}\)
