初二下学期分式加减应用求解~
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初二下学期分式加减应用求解~
已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……
+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
已知:(1/m) - 1/(m+1) = 1/[m(m+1)] 请你化简:1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……
+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
![初二下学期分式加减应用求解~](/uploads/image/z/16625802-66-2.jpg?t=%E5%88%9D%E4%BA%8C%E4%B8%8B%E5%AD%A6%E6%9C%9F%E5%88%86%E5%BC%8F%E5%8A%A0%E5%87%8F%E5%BA%94%E7%94%A8%E6%B1%82%E8%A7%A3%7E)
1/[(n+1)(n+2)] + 1/[n+2)(n+3)] +……+ 1/[(n+2006)(n+2007)] + 1/[(n+2007)(n+2008)]
=1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+······+1/(n+2007)-1/(n+2008)
=1/(n+1)-1/(n+2008)
=2007/(n+1)(n+2008)
=1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+······+1/(n+2007)-1/(n+2008)
=1/(n+1)-1/(n+2008)
=2007/(n+1)(n+2008)