F.1數學題

2006-12-29 3:09 am
基礎代數:
1.(2+3t-4t^2)+(5-6t^2)
2.(f+3)-(3f-4)-(5+8f)
3.(2r^3-6-4r^2)-(3r+6r^2-5)
p.s:^2等於二次,^3等於三次
以上三條請列明步驟和答案
4.常笑原有$5t,用了$(3t-5)和$(7-2t)後,他還剩多少錢??
請列明式,答案和步驟
方程式:
x+6x+4+11-2x=180
x等於多少??
請列明步驟和答案

回答 (3)

2006-12-29 3:24 am
✔ 最佳答案
1. (2+3t-4t^2)+(5-6t^2)

= 2+3t-4t^2 +5-6t^2

= 2+5+3t-4t^2-6t^2 (呢度我只係 group 埋佢等你易睇d~)

= 7+3t-10t^2

2. (f+3)-(3f-4)-(5+8f)

= f+3-3f + 4 -5 - 8f

= 2 - 10f

3. (2r^3-6-4r^2)-(3r+6r^2-5)

= 2r^3 - 6 - 4r^2 - 3r - 6r^2 +5

= 2r^3 -10r^2 -3r -1

4. 他還餘 : 5t - (3t-5) - (7-2t)

= 5t - 3t + 5 - 7 + 2t

= 4t - 2

x+6x+4+11-2x=180

5x + 15 = 180

5x+15-15=180-15

5x = 165

5x / 5 = 165 / 5

x = 33
參考: me
2006-12-29 4:18 am
1. (2+3t-4t^2)+(5-6t^2)
= 2+3t-4t^2+5-6t^2
= 2+5+3t-4t^2-6t^2
= 7+3t-10t^2

2. (f+3)-(3f-4)-(5+8f)
= f+3-3f+4-5-8f
= 2-10f

3. (2r^3-6-4r^2)-(3r+6r^2-5)
= 2r^3-6-4r^2-3r-6r^2+5
= 2r^3-10r^2-3r-1

4. 5t-(3t-5)-(7-2t)
= 5t-3t+5-7+2t
= 5t-3t+2t+5-7
= 4t-2
他還剩$(4t-2)

方程式:
x+6x+4+11-2x=180
5x+15=180
5x+15-15=180-15
5x=165
5x/5=165/5
x=33
參考: myself
2006-12-29 4:01 am
1: (2+3t-4t^2)+(5-6t^2)
  =2+3t-4t^2+5-6t^2
  =7+3t-10t^2

2: (f+3)-(3f-4)-(5+8f)
  =f+3-3f+4-5+8f
  =6f+2

3: (2r^3-6-4r^2)-(3r+6r^2-5)
  =2r^3-6-4r^2-3r-6r^2+5
  =2r^3-10r^2-3r-1

4:He has:$〔5t-(3t-5)-(7-2t)〕
        =$〔5t-3t+5-7+2t〕
        =$〔4t-2〕

5:x+6x+4+11-2x=180
         5x+15=180
            5x=165
             x=53 

2006-12-28 20:02:20 補充:
2: (f+3)-(3f-4)-(5+8f)  =f+3-3f+4-5-8f  =2-10f
參考: MYSELF


收錄日期: 2021-04-23 19:19:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061228000051KK01892

檢視 Wayback Machine 備份