3z-2分之7z+2-3z分之z+4
要有式啊！
好急
maths questions?
2010-10-10 10:08 pm
回答 (3)
2010-10-10 10:43 pm
✔ 最佳答案
7z/(3z - 2) + (z + 4)/(2 - 3z)= 7z/(3z - 2) - (z + 4)/(3z - 2)
= [7z - (z + 4)] / (3z - 2)
= (7z - z - 4) / (3z - 2)
= (6z - 4) / (3z - 2)
= 2(3z - 2) / (3z - 2)
= 2
參考: 土扁
2010-10-10 10:52 pm
7z/(3z-2)+[(z+4)/2-3z]
=7z/(3z-2)+[(z+4)/-(3z-2)]
=7z/(3z-2)-[(z+4)/3z-2]
=(7z-z-4)/(3z-2)
=(6z-4)/(3z-2)
=2(3z-2)/(3z-2)
=2
=7z/(3z-2)+[(z+4)/-(3z-2)]
=7z/(3z-2)-[(z+4)/3z-2]
=(7z-z-4)/(3z-2)
=(6z-4)/(3z-2)
=2(3z-2)/(3z-2)
=2
參考: me
2010-10-10 10:12 pm
no..........................................
收錄日期: 2021-04-13 17:34:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101010000051KK00657