问题描述: (1/2010-1)*(1/2009-1)*(1/2008-1)*...*(1/1001-1)*(1/1000-1) 1个回答 分类:数学 2014-12-03 问题解答: 我来补答 (1/2010-1)*(1/2009-1)*(1/2008-1)*...*(1/1001-1)*(1/1000-1)=(-1)的1011次方*(1-1/2010)*(1-1/2009)*(1-1/2008)*...*(1-1/1001)*(1-1/1000)=-(2009/2010)*(2008/2009)*(2007/2008)*...*(1000/1001)*(999/1000)=-999/2010 展开全文阅读