求值3x^2y-[2xyz-(2xyz-x^2y)-4x^2y]-2xyz其中x=-2,y=-3,z=1

ivfhlcdr2022-10-04 11:39:542条回答

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?

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粉红ing 共回答了18个问题 | 采纳率100%
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
1年前
紫晶杨梅 共回答了92个问题 | 采纳率
x^4+x^3+2x^2+x+1
=x^4+x^3+x^2+x^2+x+1
=x^2(x^2+x+1)+x^2+x+1
=(x^2+1)(x^2+x+1)
-1-2x-x^2+y^2
=y^2-(1+2x+x^2)
=y^2-(x+1)^2
=(y+x+1)(y-x-1)
x^2+ax^2+x+ax-1-a
=x^2(a+1)+x(a+1)-(a+1)
=(a+1)(x^2+x-1)
1年前

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