1/(a-c)(a-b) + 1/(b-a)(b-c) + 1/(c-a)(c-b)
Simplify it
give me the full step
F.2 Math question
2009-12-23 6:07 am
回答 (1)
2009-12-23 6:37 am
✔ 最佳答案
1/(a-c)(a-b) + 1/(b-a)(b-c) + 1/(c-a)(c-b)= 1/(a-c)(a-b) - 1/(a-b)(b-c) + 1/(a-c)(b-c)
通分母
= (b-c)/(a-c)(a-b)(b-c) - (a-c)/(a-c)(a-b)(b-c) + (a-b)/(a-c)(a-b)(b-c)
= (b - c - a + c + a - b)/(a-c)(a-b)(b-c)
= 0
收錄日期: 2021-04-23 23:22:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091222000051KK01569