用简便算法计算:(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 07:57:31
用简便算法计算:(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)

用简便算法计算:(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)
用简便算法计算:(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)
(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)

用简便算法计算:(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)
(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)-(1-1/2-1/3-1/4-…-1/2010)×(1/2+1/3+1/4+…+1/2010)
=(1-1/2-1/3-1/4-…-1/2010)×[(1/2+1/3+1/4+…+1/2010)-(1/2+1/3+1/4+…+1/2010)]
=(1-1/2-1/3-1/4-…-1/2010)×0
=0

这题是不是写错了 ?没错的话原式=0

这需要算吗?