计算[-(-2x^2y)^3]^2·(-3x^3y^2)
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解决时间 2021-03-12 09:04
- 提问者网友:锁深秋
- 2021-03-12 03:02
计算[-(-2x^2y)^3]^2·(-3x^3y^2)
最佳答案
- 五星知识达人网友:青尢
- 2021-03-12 04:12
(1) [-(-2x^2 y)^3]^2·(-3x^3 y^2)
=[-(-8x^6y^3)]^2*(-3x^3y^2)
=64x^12y^6*(-3x^3y^2)
=-192x^15y^8.
(2)[x^(2n-1)]^2·(x^2)^(3-2n)·[-x^(2n-3)]
=x^(4n-2)*x^(6-4n)*[-x^(2n-3)]
=-x^(4n-2+6-4n+2n-3)
=-x^(2n+1).
=[-(-8x^6y^3)]^2*(-3x^3y^2)
=64x^12y^6*(-3x^3y^2)
=-192x^15y^8.
(2)[x^(2n-1)]^2·(x^2)^(3-2n)·[-x^(2n-3)]
=x^(4n-2)*x^(6-4n)*[-x^(2n-3)]
=-x^(4n-2+6-4n+2n-3)
=-x^(2n+1).
全部回答
- 1楼网友:骨子里都是戏
- 2021-03-12 04:17
[(x+y/x-y)^2*(2y-2x/3x+3y)]-[(x^2/x^2-y^2)/(x/y)]={[(x+y)/(x-y)]^2*(2y-2x)/(3x+3y)}-[x^2/(x^2-y^2)*y/x]=-{[(x+y)/(x-y)]^2*2(x-y)/3(x+y)}-[x^2/(x^2-y^2)*y/x]=-(x+y)/(x-y)-[x/(x^2-y^2)*y]=-(x+y)/(x-y)-[xy/(x^2-y^2)]=-(x+y)^2/(x^2-y^2)-[xy/(x^2-y^2)]=-[(x+y)^2+xy]/(x^2-y^2)]=-(x^2+y^2+2xy+xy)/(x^2-y^2)=-(x^2+y^2+3xy)/(x^2-y^2)
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