最後一條數 唔該幫下手
回答 (3)
2007-10-22 7:55 am
✔ 最佳答案
21(n-3m/7)(m+7n/3)= 21(mn + 7n^2/3 - 3m^2/7 - mn)
= 21 (7n^2/3 - 3m^2/7)
= 21 (7n^2/3) - 21(3m^2/7)
= 7(7n^2) - 3(3m^2)
= 49n^2 - 9m^2
^ 指次方, ^2是指2次方
2007-10-22 00:44:15 補充：
有一條式給你很有參考作用的: (a+b) (c-d) = ac - ad + bc - bd希望對你有幫助...
參考: 自己
2007-10-22 7:51 am
21(n-3m/7)(m+7n/3)
=21 [ mn + (7/3)n^2 - (3/7)m^2 - (21/21)mn ]
=21 [ (7/3)n^2 - (3/7)m^2 ]
=49n^2 - 9m^2
=21 [ mn + (7/3)n^2 - (3/7)m^2 - (21/21)mn ]
=21 [ (7/3)n^2 - (3/7)m^2 ]
=49n^2 - 9m^2
參考: me
2007-10-22 7:38 am
(n-3m/7)(m+7n/3)
=21(nm+(7n^2)/3-(3m^2)/7+21mn/21)
=21nm+49n^2-9m^2+21mn
=42mn+49n^2-9m^2
2007-10-21 23:39:21 補充：
sorry抄題目果度少左21兩個字
=21(nm+(7n^2)/3-(3m^2)/7+21mn/21)
=21nm+49n^2-9m^2+21mn
=42mn+49n^2-9m^2
2007-10-21 23:39:21 補充：
sorry抄題目果度少左21兩個字
收錄日期: 2021-04-13 16:07:54
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071021000051KK06027