can you help me to solve this maths question?
回答 (3)
2006-12-26 11:54 pm
✔ 最佳答案
由於括號內有負號，因此要先處理括號內，先通分母。方法一
(a-1﹣b-1)-1
= (1/a﹣1/b)-1
= (b/ab﹣a/ab)-1
= (b﹣a/ab)-1
= (b﹣a)-1 / (ab)-1
= ab / b﹣a
應用了負指數 x-1 = 1/x
方法二
(a-1﹣b-1)-1
= 1 / (a-1﹣b-1)1
= 1 / (1/a﹣1/b)1
= 1 / (b/ab﹣a/ab)1
= 1 / (b﹣a/ab)
= ab / b﹣a
2006-12-26 11:07 pm
(a^-1 - b^-1)^-1
=(1/a - 1/b)^-1
=[(b-a)/ab]^-1
=ab/(b-a)
=(1/a - 1/b)^-1
=[(b-a)/ab]^-1
=ab/(b-a)
2006-12-26 11:07 pm
(a^-1 - b^-1- )^-1
=(1/a - 1/b)^-1
= (b/ab - a/ab)^-1
= ((b-a)/ab)^-1
= 1 / ((b-a)/ab)
= ab / (b-a)
=(1/a - 1/b)^-1
= (b/ab - a/ab)^-1
= ((b-a)/ab)^-1
= 1 / ((b-a)/ab)
= ab / (b-a)
收錄日期: 2021-04-12 22:00:03
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061226000051KK02180