初二下学期分式加减应用求解~
问题描述：
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)]
答
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)