\(A=2012-\left(2x^2+5y^2-2xy-4x-4y\right)\\ \)
Hệ số lẻ quá:
B=2A đặt 2x=z
\(B=m-\left(z^2+10y^2-2yz-4z-8y\right)\)
\(B=m-\left[\left(z-y-2\right)^2+9y^2-12y-4\right]\)
\(B=m-\left[\left(z-y-2\right)^2+\left(t-2\right)^2-4-4\right]\)
\(B=\left(m-8\right)-\left(z-y-2\right)^2-\left(t-2\right)^2\)
\(A_{min}=\frac{2.2012-8}{2}=2008\)đạt tại \(\orbr{\begin{cases}t-2=0=>y=\frac{2}{3}\\z-y-2=0\Rightarrow x=\frac{4}{3}\end{cases}}\)