E=x^2+2y^2-2xy+2x-10y
=x2+y2-2xy+y2-8y+16+2x-2y-16
=(x-y)2+(y-4)2+2.(x-y)-16
=(x-y)2+2(x-y)+1+(y-4)2-17
=(x-y+1)2+(y-4)2-17 \(\ge\)-17
Dấu "=" xảy ra khi: y=4; x=3
Vậy GTNN của E là -17 tại x=3;y=4
\(E=x^2+2y^2-2xy+2x-10y\)
\(=\left(x^2-2xy+2x\right)+2y^2-10y\)
\(=x^2-2x\left(y+1\right)+2y^2-10y\)
\(=x^2-2x\left(y+1\right)+\left(y-1\right)^2+2y^2-10y-\left(y-1\right)^2\)
\(=\left[x-\left(y-1\right)\right]^2+2y^2-10y-y^2+2y-1\)
\(=\left(x-y+1\right)^2+y^2-8y-1=\left(x-y+1\right)^2+\left(y^2-2.y.4+16\right)-17\)
\(=\left(x-y+1\right)^2+\left(y-4\right)^2-17\)
Vì \(\left(x-y+1\right)^2\ge0;\left(y-4\right)^2\ge0=>E=\left(x-y+1\right)^2+\left(y-4\right)^2-17\ge-17\) (với mọi x;y)
Dấu "=" xảy ra \(< =>\hept{\begin{cases}\left(x-y+1\right)^2=0\\\left(y-4\right)^2=0\end{cases}< =>\hept{\begin{cases}x-y=-1\\y=4\end{cases}}< =>x=3;y=4}\)
Vậy minE=-17 khi x=3;y=4
