计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×5/6(y-x)^2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/08 21:39:38
计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×5/6(y-x)^2
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
再问: 如果....计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2呢
再答: (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
再问: (x^2-y^2)^5岔开,我就采纳你
再答: (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
再问: 呵呵 、谢谢
再答: 不客气,谢谢采纳支持,祝你学习进步
(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
再问: 如果....计算:(x+y)^2×(-x-y)^3×[-3/4(x-y)^3]×6/5(y-x)^2呢
再答: (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
再问: (x^2-y^2)^5岔开,我就采纳你
再答: (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
再问: 呵呵 、谢谢
再答: 不客气,谢谢采纳支持,祝你学习进步
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