nixiang915 发表于 2009-11-6 06:26:55

帮我做一道题。。。谢谢

本帖最后由 nixiang915 于 2009-11-5 17:39 编辑

请在10小时以内帮我解答这个数学问题。

因式分解：
(s/r)^12-(s/r)^6

更新： 我好像草草的作出来了， 这是对的吗？
设：s/r=a

(a)^12-(a)^6
=(a^4)^3-(a^2)^3
=(9a^4-a^2)(a^8+a^6=a^4)
=a^2[(a^2-1)(6^6+a^4+a^2)]
=a^2[(a-1)(a+1)a^2(a^4+a^2+1)]

这是对的吗？
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蓝冰 发表于 2009-11-6 09:35:37

s/r=a
明显过程有问题：

(a)^12-(a)^6
=(a)^6[(a)^3-1][(a)^3+1]
=(a)^6[(a)-1][(a)^2+a+1][(a)+1][(a)^2-a+1]

nixiang915 发表于 2009-11-6 09:53:35

请问是怎么从第二步到第三步的？

J-Cent 发表于 2009-11-6 16:31:07

...完了完了...看不懂了...火星了, 我这商科学得

蓝冰 发表于 2009-11-6 17:08:04

你的2-3就是我的2-3啊！
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查看完整版本: 帮我做一道题。。。谢谢