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:2024/07/05 12:51:35
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)
x){>eu OnS_[d3C]C}# 6bG @|C{:OieD( ůFlT?;p/=2!$cF 1 lg\4ٌOvɧm3_NdW=mZ

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?