求1/x(x+3)+1/（x+3)(x+6)+...+1/(x+27)(x+30)

我想应该是这样的~
1/x(x+3)+1/（x+3)(x+6)+...+1/(x+27)(x+30)
=1/3[1/x/-1/(x+3)]+1/3[1/(x+3)-1/(x+6)]+1/3[1/(x+5)-1/(x+7)]...+1/3[1/(x+27)-1/(x+30)]
=1/3[1/x-1/(x+3)+1/x+3/-1/(x+6)+....]
=1/3[1/x-1/(x+30)]
=10/x(x+30)
..希望能帮到你

因为1/x-1/(x+3)=2/x(x+3)
故1/x(x+3)=1/2[1/x-1/(x+3)]
故原算式=1/2[1/x-1/(x+3)]+1/(x+3)-1/(x+6)]+...
=1/2[1/x-1/(x+30)]

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

方法：每项拆开后两两消去。
1/x(x+3)+1/（x+3)(x+6)+...+1/(x+27)(x+30)
= [3/x(x+3)+3/（x+3)(x+6)+...+3/(x+27)(x+30)]/3
= [(x+3-x)/x(x+3)+(x+6-x-3)/（x+3)(x+6)+...+(x+30-x-27)/(x+27)(x+30)]/3
= [(1/x-1/(x+3))+(1/(x+3)-1/(x+6))+...+(1/(x+27)-1/(x+30))]/3
= [1/x-1/(x+30)]/3
= 30/(3x(x+30)
= 10/(x(x+30))