Để pt có nghiệm duy nhất thì: $\frac{m}{-2}\neq\frac{-2}{m}$
⇔ $m\neq±2$
$\left \{ {{mx-2y=3} \atop {-2x+my=1-m}} \right.$
⇔ $\left \{ {{2mx-4y=6} \atop {-2mx+m²y=m-m²}} \right.$
Cộng vế theo vế, ta có: $-2mx+m²y+2mx-4y=m-m²+6$
⇔ $m²y-4y=m-m²+6$
⇔ $y.(m-2).(m+2)=(m-3).(m+2)$
⇔ $y=\frac{m-3}{m-2}$
⇒ $x=\frac{my+m-1}{2}=\frac{\frac{m²-3m}{m-2}+m-1}{2}=\frac{2m²-6m+2}{2.(m-2)}$
= $\frac{m²-3m+1}{m-2}$
Vì $x+y=1$
⇔ $\frac{m²-3m+1}{m-2}+\frac{m-3}{m-2}=1$
⇔ $\frac{m²-3m+1+m-3}{m-2}=1$
⇔ $m²-2m-2=m-2$
⇔ $m²-3m=0$
⇔ \(\left[ \begin{array}{l}m=0\\m=3\end{array} \right.\)
