请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b

来源：百度知道 编辑：UC知道 时间：2024/06/10 17:51:47
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)]

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+2008)]

1/[x(x+3)]+1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
=(1/3){3/[x(x+3)]+3/[(x+3)(x+6)]+……+3/[(x+27)(x+30)]}
=(1/3)[1/x-1/(x+3)+1/(x+3)-1/(x+6)+……+1/(x+27)-1/(x+30)]
=(1/3)[1/x-1/(x+30)]
=10/[x(x+30)]

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+2008)

1/[x(x+3)]+1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
=[1/x-1/(x+3)/3+[1/(x+3)-1/(x+6)]/3+...+[1/(x+27)-1/(x+30)]/3
=[1/x-1/(x+30)]/3
=[30/x(x+30)]/3
=10/x(x+30)