1.
Đặt : đẳng thức đó =t
Ta có: $\left \{ {{x+y=2017t} \atop {x-y=2019t}} \right.$ ⇔$\left \{ {{2x=4036t} \atop {x+y=2017t}} \right.$
⇔$\left \{ {{x=2018t} \atop {y=-t}} \right.$
⇒$\frac{xy}{2018}=t$ ⇔(2018t)(-t)=2018t⇔t=-1
⇒$\left \{ {{x=-2018} \atop {y=1}} \right.$
2.$\frac{x}{a+2b+c}$ =$\frac{y}{2a+b-c}$ =$\frac{z}{4a-4b+c}$ =t
Ta có: t=$\frac{x+2y+z}{(a+2b+c)+2(2a+b-c)+(4a-4b+c)}$ =$\frac{x+2y+z}{9a}$ (1)
t=$\frac{2x+y-z}{2(a+2b+c)+(2a+b-c)-(4a-4b+c)}$ =$\frac{2x+y-z}{9b}$ (2)
t=$\frac{4x-4y+z}{4(a+2b+c)-4(2a+b-c)+(4a-4b+c)}$ =$\frac{4x-4y+z}{9c}$ (3)
Từ (1),(2),(3) ta có: $\frac{x+2y+z}{9a}$=$\frac{2x+y-z}{9b}$ =$\frac{4x-4y+z}{9c}$ =t
⇔ (Đpcm)