\(\left(x+y+3\right)^2=1-y^2\)
Ta thấy \(1-y^2\le1\) do \(y^2\ge0\forall y\)
Suy ra \( \left(x+y+3\right)^2\le1\Rightarrow\left|x+y+3\right|\le1\Rightarrow-1\le x+y+3\le1\)
\(\Rightarrow2012\le x+y+2016\le2014\)
\(Min_{\left(B\right)}=2012\Leftrightarrow x=-4;y=0\)
\(Max_{\left(B\right)}=2014\Leftrightarrow x=-2;y=0\)
Chúc bạn học tốt !!!