求值3x^2y-[2xyz-(2xyz-x^2y)-4x^2y]-2xyz其中x=-2,y=-3,z=1
问题描述:
求值3x^2y-[2xyz-(2xyz-x^2y)-4x^2y]-2xyz其中x=-2,y=-3,z=1
我算到一半,怎么会有0?
答
3x^2y-[2xyz-(2xyz-x^2y)-4x^2y]-2xyz=3x^2y-2xyz+(2xyz-x^2y)+4x^2y-2xyz=7x^2y-4xyz+(2xyz-x^2y)=7x^2y-4xyz+2xyz-x^2y=6x^2y-2xyz=6×(-2)^2×(-3)-2×(-2)×(-3)×1=-72-12=-84