(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011-1)=
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(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011-1)=(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011
(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011-1)=
(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011-1)=
(1/1000-1)(1/1001-1)(1/1002-1).(1/2010-1)(1/2011-1)=
原式分母从1000~2011一共有1012项,且每一项都是负数,所以乘积为正,我直接把每项取相反数结果不变.所以,原式=999/1000乘1000/1001乘1001/1002.乘2009/2010乘2010/2011
找规律,可看出从1000~2010都出现过2次,分别在分母和后一项的分子上,可以通分消去,只剩下999和2011两项,故原式=999/2011
2220
(1/2000-1)*(1/1999-1)*…*(1/1001-1)*(1/1000-1)
(1/1998-1)(1/1997-1)(1/1996-1)...(1/1001-1)(1/1000-1)
(2003/1-1)×(2002/1-1)×(2001/1-1)×.×(1001/1-1)×(1000/1-1)=?
(1998/1-1)(1997/1-1)(1996/1-1).(1001/1-1)(1000/1-1).
(1998/1-1)*(1997/1-1)*(1996/1-1)*.*(1001/1-1)*(1000/1-1).
(1-1/2006)*(1-1/2005)*(1-1/2004).(1-1/1001)*(1-1/1000)
1/(2×1001)+1/(3×1001)-1/(4×1001)+1/(6×1001)-1/(8×1001)
(1/2008-1)*(1/2007-1)*(2006-1).(1/1001-1)*(1/1000-1)
(1/1998-1)(1/1997-1)(1/1996-1)...(1/1001-1)(1/1000-1)
计算 ( 1+1/2)*(1-1/3)*(1+1/4)*(1-1/5)*.*(1+1/1000)*(1-1/1001)
(2010/1-1)x(2009/1-1)x(2008/1-1).(1001/1-1)x(1000/1-1)=?
(2010/1-1)x(2009/1-1)x(2008/1-1).(1001/1-1)x(1000/1-1)=?
计算:(1/2014-1)*(1/2013-1)*(1/2012-1)*.*(1/1001-1)*(1/1000-1)
计算:(1+1/2)×(1-1/3)×(1+1/4)×(1-1/5)×.×(1+1/1000)×(1-1/1001)
1001+1002×1000/1001×1002-1简算
|1001/1-1000/1|+|1002/1-1001/1|=|1001/1-1000/1|+|1002/1-1001/1|
求(1/2001-1)*(1/2000-1)*(1/1999-1)*...*(1/1001-1)*(1/1000-1)的值怎么做?
(1/2012-1)*(1/2011-1)*(1/2010-1)*……*(1/1001-1)*(1/1000-1)