第1題
10000-9999+9998-9997+.....+4-3+2-1=?
第2題
10-9+8-7+6-5+4-3+2-1=?
兩條數學題 急急急!!!!!! 請幫忙
2007-08-14 10:21 pm
回答 (7)
2007-08-14 10:29 pm
✔ 最佳答案
1. 10000-9999+9998-9997+.....+4-3+2-1= (10000-9999)+(9998-9997)+......(4-3)+(2-1)
= 1+1+.....+1+1
= 1 * 5000
= 5000
2. 10-9+8-7+6-5+4-3+2-1
= (10-9)+(8-7)+(6-5)+(4-3)+(2-1)
= 1+1+1+1+1
= 5
2007-08-15 10:32 pm
1. 10000-9999+9998-9997+.....+4-3+2-1
= (10000-9999)+(9998-9997)+......(4-3)+(2-1)
= 1+1+.....+1+1
= 1 * 5000
= 5000
2. 10-9+8-7+6-5+4-3+2-1
= (10-9)+(8-7)+(6-5)+(4-3)+(2-1)
= 1+1+1+1+1
= 5
= (10000-9999)+(9998-9997)+......(4-3)+(2-1)
= 1+1+.....+1+1
= 1 * 5000
= 5000
2. 10-9+8-7+6-5+4-3+2-1
= (10-9)+(8-7)+(6-5)+(4-3)+(2-1)
= 1+1+1+1+1
= 5
2007-08-14 11:03 pm
1. 10000-9999+9998-9997+......+4-3+2-1
=(10000-9999)+(9998-9997)+......+(4-3)+(2-1)
=1+1+.....+1+1
=5000
2. 10-9+8-7+6-5+4-3+2-1
=(10-9)+(8-7)+(6-5)+(4-3)+(2-1)
=1+1+1+1+1
=5
In fact, we only need to use brackets to separate them into different groups and we can find out the answers of the two questions above easily.
*Be careful that a-b-c is not equal to a-(b-c).
=(10000-9999)+(9998-9997)+......+(4-3)+(2-1)
=1+1+.....+1+1
=5000
2. 10-9+8-7+6-5+4-3+2-1
=(10-9)+(8-7)+(6-5)+(4-3)+(2-1)
=1+1+1+1+1
=5
In fact, we only need to use brackets to separate them into different groups and we can find out the answers of the two questions above easily.
*Be careful that a-b-c is not equal to a-(b-c).
2007-08-14 10:38 pm
第1題:5000
第2題:5
第2題:5
2007-08-14 10:37 pm
第一題係
((10000/2 * (10000+2)) / 2) -
(((9999+1)/2 * (9999+1)) / 2)
=5000
((10000/2 * (10000+2)) / 2) -
(((9999+1)/2 * (9999+1)) / 2)
=5000
參考: 自己
2007-08-14 10:28 pm
1. 5000
2. 5
2. 5
2007-08-14 10:25 pm
第2題係5
收錄日期: 2021-04-18 22:57:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070814000051KK02567