有數學唔識做??

2007-08-21 2:42 am
化簡
1.(12a-b-7c)-7a+8c-(3a-7b-4c)
2.(5b2次方+6b-3)-(3b2次方-2b+4)
展開
例:2a(a-3) =2axa+2ax(-3) =2a2次方
3.3c-2=13
4.4(n-5)=12
要有過程

回答 (6)

2007-08-21 2:55 am
✔ 最佳答案
1. (12a-b-7c)-7a+8c-(3a-7b-4c)
=12a-b-7c -7a+8c-3a+7b+4c
=2a-6b+5c -->注意:括號唔要時,小心負負得正or負正得負

2.(5b^2+6b-3)-(3b^2-2b+4)
=5b^2+6b-3-3b^2+2b-4
=2b^2+8b-7-->注意:括號唔要時,小心負負得正or負正得負

3.好似係方程wo,你係唔係想解方程呢 ?
3c-2=13
3c=15
c=5

4. 4(n-5)=12
4n-20=12
4n=32
n=8

如果係展開,應該得一條式,沒有"=" xx你在貼多一次…
2007-08-21 3:16 am
化簡

1)
(12a-b-7c)-7a+8c-(3a-7b-4c)

=12a-b-7c-7a+8c-3a+7b+4c

=12a-7a-3a-b+7b-7c+8c+4c

=2a+6b+5c

2)
(5b2次方+6b-3)-(3b2次方-2b+4)

=2b^2+6b-3-3b^2+2b+4

=2b^2-3b^2+6b+2b+4

=8b-b^2+4

嚴禁抄錄答案
2007-08-21 3:03 am
1. (12a-b-7c)-7a+8c-(3a-7b-4c)
=12a-b-7c-7a+8c-3a+7b+4c
=12a-7a-3a-b+7b-7c+8c+4c
=2a-6b+5c
=======
2. (5b2次方+6b-3)-(3b2次方-2b+4)
=5b2次方+6b-3-3b2次方+2b-4
=5b2次方-3b2次方+6b+2b-3-4
=2b2次方+8b-7
==========
3. 3c-2=13
3c-2+2=13+2
3c=15
3c/3=15/3
c=5
==
4. 4(n-5)=12
4n-20=12
4n-20+20=12+20
4n=32
4n/4=32/4
n=8
==
參考: me
2007-08-21 2:54 am
1.(12a-b-7c)-7a+8c-(3a-7b-4c)
=12a-b-7c-7a+8c-3a+7b+4c
=2a+6b+5c

2.(5b2次方+6b-3)-(3b2次方-2b+4)
=(25b的2次方+6b-3)-(9b的2次方-2b+4)
=25b的2次方+6b-3-9b的2次方+2b-4
=16b的2次方+8b-7

3.3c-2=13
3c=11
c=11/3

4.4(n-5)=12
n-5 = 3
n = 8

2007-08-20 18:55:08 補充:
3. 3c-2=13 3c = 15 c= 5
參考: me
2007-08-21 2:51 am
1.12a-b-7c-7a+8c-3a+7b+4c
=12a-7a-3a-b+7b-7c+8c+4c
=2a+6b+5c

2.5b^2+6b-3-3b^2+2b+4
=5b^2-3b^2+6b+2b-3+4
2b^2+8b+1

3.3c-2=13
3c=13+2
3c=15
c=5

4.4n-4(5)=12
4n-20=12
4n=12+20
4n=32
n=8

2007-08-20 20:59:49 補充:
2.(5b2次方 6b-3)-(3b2次方-2b 4)若是[(5b)^2 6b-3]-[(3b)^2-2b 4],則應為:25b^2 6b-3-9b^2 2b-4=25b^2-9b^2 6b 2b-3-4=16b^2 8b-7

2007-08-20 21:01:03 補充:
另一答案(正文有錯漏):2.5b^2 6b-3-3b^2 2b-4=5b^2-3b^2 6b 2b-3-42b^2 8b-7
2007-08-21 2:50 am
1) (12a-b-7c)-7a+8c-(3a-7b-4c)
=12a-b-7c-7a+8c-3a+7b+4c
=12a-7a-3a-b+7b-7c+8c+4c
=2a+6b+5c

2) (5b2次方+6b-3)-(3b2次方-2b+4)
=2b^2+6b-3-3b^2+2b+4
=2b^2-3b^2+6b+2b+4
=8b-b^2+4

第3、4題不太明要如何展開@@?
只能把未知數算出.."
參考: 自己


收錄日期: 2021-04-13 13:01:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070820000051KK04326

檢視 Wayback Machine 備份