數學題:我要方法(快,簡單)

2007-05-15 4:08 am
數學題:我要方法(快,簡單)
1-3+5-7+9-11......+2001-2003+2005=?

1989*17891789-1789*18991899=?

123123123/193193193+456456456/193193193=?

回答 (2)

2007-05-15 5:10 am
✔ 最佳答案
1. 項數=(2005-1)/÷2+1=1003
把 (1-3)、(5-7) 為視為一組,組別數目=(1003-1)÷2=501
即是 (-2)×501=(-1002)
答案=2005+(-1002)=1003

2.1989*17891789-1789*18991899
=1989×1789×10001-1789×1899×10001
=0

2.123123123/193193193+456456456/193193193
=123123123/193193193+123123123×3/193193193
=123123123×4/193193193
=123×1001001×4/193×1001001
=123×4/193
=492/193

2007-05-14 21:13:00 補充:
第三題錯了 ==3. 123123123/193193193 456456456/193193193=123×1001001/193×1001001 456×1001001/193×1001001=123/193 456/193=(123 456)/193=579/193=3

2007-05-14 21:15:03 補充:
沒了+號==123123123/193193193+456456456/193193193=123×1001001/193×1001001+456×1001001/193×1001001=123/193+456/193=(123+456)/193=579/193=3
參考: 個人計算
2007-05-19 3:17 am
good


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https://hk.answers.yahoo.com/question/index?qid=20070514000051KK04056

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