Câu 11:
\(x^2+2y^2+2xy+2x+6y+1=0\\ \Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)+1+y^2+4y+4-4=0\\ \Leftrightarrow\left(x+y+1\right)^2+\left(y+2\right)^2=4\\ \Leftrightarrow4-\left(x+y+1\right)^2=\left(y+2\right)^2\ge0\\ \Leftrightarrow\left(x+y+1\right)^2\le4\\ \Leftrightarrow-2\le x+y+1\le2\\ \Leftrightarrow A=x+y+1+2018\le2+2018=2020\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y+1=2\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)
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