simpify the following:
圖片參考：http://imgcld.yimg.com/8/n/HA00407912/o/701204140074813873394070.jpg
this first one is : 1 over a(a-b)(a-c)
SIMPIFY
2012-04-15 4:49 am
回答 (1)
2012-04-15 8:39 am
✔ 最佳答案
SIMIPY 1/[a(a－b)(a－c)+1/[b(b－c)(b－a)+1/[c(c－a)(c－b)] Sol
p=1/[a(a－b)(a－c)]+1/[b(b－c)(b－a)]+1/[c(c－a)(c－b)]
pabc(a－b)(b－c)(c－a)
=bc(c－b)+ac(a－c)+ab(b－a)
=bc^2－b^2c+a^2c－ac^2+ab^2－a^2b
=(bc^2－ac^2)－(b^2c－a^2c)+(ab^2－a^2b)
=c^2(b－a)－c(b^2－a^2)+ab(b－a)
=c^2(b－a)－c(b－a)(b+a)+ab(b－a)
=(b－a)(c^2－bc－ac+ab)
=(b－a)[c(c－b)－a(c－b)]
=(b－a)(c－a)(c－b)
=(a－b)(b－c)(c－a)
So
pabc=1
p=1/(abc)
收錄日期: 2021-04-13 18:38:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120414000051KK00748