b,
1, Vì $\dfrac{m}{-1}\neq\dfrac{-1}{-m}$
⇒ Phương trình luôn có nghiệm với $∀m$
2, Ta có: $\left \{ {{mx-y=12} \atop {-x-my=-3}} \right.$
⇔ $\left \{ {{m²x-my=12m} \atop {x+my=3}} \right.$
⇔ $\left \{ {{m²x-my+x+my=12m+3} \atop {y=mx-2}} \right.$
⇔ $\left \{ {{(m²+1)x=12m+3} \atop {y=mx-2}} \right.$
⇔ $\left \{ {{x=\dfrac{12m+3}{m²+1}} \atop {y=\dfrac{10m²+3m-2}{m²+1}}} \right.$
Ta có: $2x+y=0$
⇔ $2.\dfrac{12m+3}{m²+1}+\dfrac{10m²+3m-2}{m²+1}=0$
⇔ $\dfrac{2.(12m+3)+10m²+3m-2}{m²+1}=0$
⇒ $2.(12m+3)+10m²+3m-2=0$
⇔ $10m²+27m+4=0$
Ta có: $Δ=27²-4.4.10=569$
⇒ $m_{1}=\dfrac{-27+\sqrt{569}}{2.10}$
và $m_{2}=\dfrac{-27-\sqrt{569}}{2.10}$
