Question

计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×5/6(y-x)^2

有过程给加分，没过程不采纳

Answer 1

答：
(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×5/6(y-x)^2
=[(x+y)^2]×[-(x+y)^3]×(-3/4)×[(x-y)^3]×(5/6)×(x-y)^2
=-1×(-3/4)×(5/6)×[(x+y)^(2+3)]×[(x-y)^(3+2)]
=(5/8)×[(x+y)^5]×[(x-y)^5]
=(5/8)×[(x+y)(x-y)]^5
=(5/8)×(x^2-y^2)^5

追问

如果....计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2呢

追答

(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2
=[(x+y)^2]×[-(x+y)^3]×(-3/4)×[(x-y)^3]×(6/5)×(x-y)^2
=-1×(-3/4)×(6/5)×[(x+y)^(2+3)]×[(x-y)^(3+2)]
=(9/10)×[(x+y)^5]×[(x-y)^5]
=(9/10)×[(x+y)(x-y)]^5
=(9/10)×(x^2-y^2)^5

追问

(x^2-y^2)^5岔开，我就采纳你

追答

(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2
=[(x+y)^2]×[-(x+y)^3]×(-3/4)×[(x-y)^3]×(6/5)×(x-y)^2
=-1×(-3/4)×(6/5)×[(x+y)^(2+3)]×[(x-y)^(3+2)]
=(9/10)×[(x+y)^5]×[(x-y)^5]

最后两步不要就可以了啊

追问

呵呵 、谢谢

追答

不客气，谢谢采纳支持，祝你学习进步

Answer 2

(x+y)²(-x-y)³×[-3/4(x-y)³]×5/6(y-x)²
=(x+y)²[-(x+y)³]×[-3/4(x-y)³]×5/6(x-y)²
=[(-1)×(-3/4)×5/6][(x+y)^(2+3)][(x-y)^(3+2)]
=5/8(x+y)^5(x-y)^5

追问

如果....计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2呢

追答

(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2
=[(x+y)^2]×[-(x+y)^3]×(-3/4)×[(x-y)^3]×(6/5)×(x-y)^2
=[-1×(-3/4)×(6/5)][(x+y)^(2+3)]×[(x-y)^(3+2)]
=(9/10)[(x+y)^5][(x-y)^5]
=9/10(x+y)^5(x-y)^5

追问

就是不能采纳俩，要不我把你也采纳了

追答

采纳一个错的！