$\qquad \begin{cases}mx-y=1\ (1)\\x+my=2\ (2)\end{cases}$
Từ `(1)=>y=mx-1` thay vào $(2)$
`(2)<=>x+m(mx-1)=2`
`<=>x+m^2x-m=2`
`<=>(m^2+1)x=m+2`
`<=>x={m+2}/{m^2+1}`
`=>y=mx-1=m. {m+2}/{m^2+1} -1`
`<=>y={m^2+2m-(m^2+1)}/{m^2+1}={2m-1}/{m^2+1}`
Ta có: `x+y=1`
`<=>{m+2}/{m^2+1}+{2m-1}/{m^2+1}=1`
`<=>m+2+2m-1=m^2+1`
`<=>m^2-3m=0`
`<=>m(m-3)=0`
$⇔\left[\begin{array}{l}m=0\\m-3=0\end{array}\right.$ $⇔\left[\begin{array}{l}m=0\\m=3\end{array}\right.$
Vậy `m\in {0;3}`
