请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.用上述规_数学解答

请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.
用上述规律,计算：
1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]
和：
1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]

人气：311 ℃　时间：2020-06-27 02:10:58

解答

1/[x(x+1)]=[(x+1)-x]/[x(x+1)]=(x+1)/[x(x+1)]-x/[x(x+1)]=1/x-1/(x+1)1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]=1/x-1/(x+1)-1/(x+1)-1/(x-2)+……+1/(x+2007)-1/(x+2008)=1/x-1/(x+2008)=2008/[x(x+20...