\(A=x^2+2y^2+2xy+2x+5y+2018\\ =\left(x^2+2xy+y^2\right)+2\left(x+y\right)+1+\left(y^2+3y+2,25\right)+2014,75\\ =\left(x+y\right)^2+2\left(x+y\right)+1+\left(y+1,5\right)^2+2014,75\\ =\left(x+y+1\right)^2+\left(y+1,5\right)^2+2014,75\)
Với mọi x ;y thì \(\left(x+y+1\right)^2\ge1\\ \left(y+1,5\right)^2\ge0\\ \Rightarrow\left(x+y+1\right)^2+\left(y+1,5\right)^2+2014,75\ge2015,75\)
Hay \(A\ge2015,75\) với mọi x;y
Để A=2014,75 thì
\(\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\\\left(y+1,5\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\\\y=-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2,5\\\\y=-1,5\end{matrix}\right.\)
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