Đáp án:
`M=x^2+2y^2+2xy-2x-6y+2015`
`M=x^2+y^2+1+2xy-2x-2y+y^2-4y+4+2010`
`M=(x^2+y^2+1+2xy-2x-2y)+(y^2-4y+4)+2010`
`M=(x+y-1)^2+(y-2)^2+2010`
Ta có: $\left\{\begin{matrix}(x+y+1)^2≥0& \\(y-2)^2≥0& \end{matrix}\right.$
`-> (x+y-1)^2+(y-2)^2>=0`
`-> (x+y-1)^2+(y-2)^2+2010>=2010`
Dấu "=" xảy ra `<=>` $\left\{\begin{matrix}(x+y+1)^2=0& \\(y-2)^2=0& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x+y+1=0& \\y-2=0& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x+y=-1& \\y=2& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x=1& \\y=2& \end{matrix}\right.$
Vậy `M_(min)=2010 <=>` $\left\{\begin{matrix}x=1& \\y=2& \end{matrix}\right.$
