\(M=x^2+6y^2+2y+36-12x-2xy\\ =\left(x^2-2xy+y^2\right)-12x+12y+36+\left(5y^2-10y+5\right)-5\\ =\left[\left(x-y\right)^2-12\left(x-y\right)+36\right]+5\left(y^2-2y+1\right)-5\\ =\left(x-y-6\right)^2+5\left(y-1\right)^2-5\)
`AA x;y` ta có `{:((x-y-6)^2>=0),(5(y-1)^2>=0):}}`
`=>(x-y-6)^2+5(y-1)^2>=0`
`=>(x-y-6)^2+5(y-1)^2-5>=0-5`
hay `M>=-5`
Dấu `" = "` xảy ra khi \(\left\{{}\begin{matrix}\left(x-y-6\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y-6=0\\y-1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-7=0\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
Vậy `Mi n M=-5` khi `x=7;y=1`
