(4+c)(5-c)
=?
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2007-11-01 1:28 am
回答 (6)
2007-11-01 4:03 am
(4+c)(5-c)
(4x5,之後4x(-c),之後c x5,之後c x c)so
=20-4c+5c-c^2
(約簡)
=20-9c-c^2
(變做descending order)
=-c^2-9c+20
(4x5,之後4x(-c),之後c x5,之後c x c)so
=20-4c+5c-c^2
(約簡)
=20-9c-c^2
(變做descending order)
=-c^2-9c+20
參考: by myself,now teaching
2007-11-01 1:36 am
(4+c)(5-c)=
4x5 + 4x(-c) + (cx5) + c(-c)=
20-4c+5c-c2=
20+c-c2
*c2 is c square
4x5 + 4x(-c) + (cx5) + c(-c)=
20-4c+5c-c2=
20+c-c2
*c2 is c square
2007-11-01 1:35 am
(4+c)(5-c)
=4*5-4c+5c-c*c
=20+9c-c*c
2007-10-31 17:38:34 補充：
=20+c-c*c
=4*5-4c+5c-c*c
=20+9c-c*c
2007-10-31 17:38:34 補充：
=20+c-c*c
2007-11-01 1:34 am
展開?
(4+c)(5-c)
=20-4c+5c-c^2
=20+c-c^2
(4+c)(5-c)
=20-4c+5c-c^2
=20+c-c^2
2007-11-01 1:34 am
(4+c)(5-c)
=(4x5+5xc)-(4xc+cxc)
=20+5c-4c-cxc
=20+c-c^2
=(4x5+5xc)-(4xc+cxc)
=20+5c-4c-cxc
=20+c-c^2
收錄日期: 2021-04-13 14:17:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071031000051KK02261