有一些有關數學的問題...

2008-12-26 11:14 pm
有些數學的問題...
1) 25u-(5u+3)(5u-3)
2) (k+1)(k+2)(k+3)
3) m/4+m/5
4) 1/p-1+1/1-p
5) 8-c/2-c+3/c-2
6) (a+b)x + (a+b)y
7)5p(x-y+z) + q(x-y+z)
Thz~

回答 (2)

2008-12-26 11:35 pm
✔ 最佳答案
以上問題想求什麼?化簡?展開?

1)
25u-(5u+3)(5u-3)
=25u-[(5u)^2-3^2]
=25u-(25u^2-9)
= -25u^2+25u+9

2)
(k+1)(k+2)(k+3)
=(k^2+3k+2)(k+3)
=k^3+3k^2+2k+3k^2+9k+6
=k^3+6k^2+11k+6

3)
m/4+m/5
=5m/20+4m/20
=(5m+4m)/20
=9m/20

4)
1/p-1+1/1-p
=1/(p-1)+1/-(p-1)
=1/(p-1)-1/(p-1)
=0
5)
8-c/2-c+3/c-2
=(8-c)/(2-c)+3/-(2-c)
=(8-c)/(2-c)-3/(2-c)
=(8-c-3)/(2-c)
=(5-c)/(2-c)

6)
(a+b)x + (a+b)y
=(a+b)(x+y)

7)
5p(x-y+z) + q(x-y+z)
=(x-y+z)(5p+q)
參考: me
2009-01-03 11:49 pm
1)25u-(5u+3)(5u-3)
=25u-5u^2-3^2
= 5u^2-25u-9
(001錯GA= =)

2)(k+1)(k+2)(k+3)
=(k^2+3k+2)(k+3)
=k^3+3k^2+2k+3k^2+9k+6
=k^3+6k^2+11k+6

3)m/4+m/5
=5m/20+4m/20
=(5m+4m)/20
=9m/20

4)1/p-1+1/1-p
=1/(p-1)+1/-(p-1)
=1/(p-1)-1/(p-1)
=0

5)8-c/2-c+3/c-2
=(8-c)/(2-c)+3/-(2-c)
=(8-c)/(2-c)-3/(2-c)
=(8-c-3)/(2-c)
=(5-c)/(2-c)

6)(a+b)x + (a+b)y
=(a+b)(x+y)

7)5p(x-y+z) + q(x-y+z)
=(x-y+z)(5p+q)

唔好信001
參考: 我=)


收錄日期: 2021-04-16 11:41:08
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081226000051KK00879

檢視 Wayback Machine 備份