+) m= -1F(x;y) = $(x- y+1)^{2} +(-x +y+1)^{2}$
=$(x-y +1)^{2} +(x-y-1 )^{2}$ đặt t= x-y+1
khi đó F(x;y) = $2t^{2} - 4t +4 =2(t -1)^{2} +2\geq2$
dấu "=' $\Leftrightarrow$ t=1 $\Leftrightarrow$ y=x
+) m $\neq$ -1
$ F\geq 0 dấu "=" \Leftrightarrow \begin{cases}x- y+1= 0\\ mx +y+m+2=0 \end{cases}$
$\Leftrightarrow \begin{cases}y=x+1 \\ mx +y +m+2=0 \end{cases}$
$\Leftrightarrow \begin{cases}x=\frac{-m -3}{m+1} \\ y=\frac{-2}{m +1} \end{cases}$