当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值

问题描述:

当(x+1)²+(y+1)²=0时,求3(2x²-xy-y²)-2(x²-2xy-3y²)的值

x=-1,y=-1;所求式子等于8