\(A=x^2+2y^2+2xy+2x-4y+2017\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+2012\)
\(=\left(x+y\right)^2+\left(x+1\right)^2+\left(y-2\right)^2+2012\ge2013\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
.........,,......
A = x2 + 2y2 + 2xy + 2x - 4y + 2017
= x2+ (2xy + 2x) + 2y2 - 4y + 2017
= x2 + 2x(y + 1) + y2 + 2y +1 + y2 - 6y + 9 + 2007
= (x + y +1)2 + (y - 3)2 + 2007 \(\ge\) 2007
Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)
Vậy Amin = 2007 khi x = -4; y = 3
