\(A=x^2-2xy+y^2+x^2+4x+2y+5\)
\(A=\left(x-y\right)^2-2\left(x-y\right)+x^2+6x+5\)
\(A=\left(x-y\right)^2-2\left(x-y\right).1+1^2+x^2+6x+4\)
\(A=\left(x-y-1\right)^2+\left(x+3\right)^2-5\)
Vậy: MinA = -5 khi............
Có thể làm theo cách này
Ta có: A = 2x2 + y2 - 2xy + 4x + 2y + 5
= (y2 - 2xy + 2y) + 2x2 + 4x + 5
= [y2 - 2y(x - 1)] + 2x2 + 4x + 5
= [y2 - 2y(x - 1) + (x - 1)2 ] - (x - 1)2 + 2x2 + 4x + 5
= (y - x +1)2 - x2 + 2x - 1 + 2x2 + 4x + 5
= (y - x +1)2 + x2 + 6x + 4
= (y - x +1)2 + x2 + 2.x.3 + 32 - 5
= (y - x +1)2 + (x +3)2 - 5
Vì (y - x +1)2 + (x +3)2 \(\ge\) 0, \(\forall\)x nên (y - x +1)2 + (x +3)2 + (-5) \(\ge\) -5, \(\forall\)x
Do đó Min A = -5 \(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\y-x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-4\end{matrix}\right.\)