求教初二数学问题。

[1/x(x+1)]+[1/(x+1)(x+2)]+[1/(x+2)(x+3)]+... ...+[1/(x+99)(x+100)]
=(1/x)-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+99)-1/(x+100)
=(1/x)-1/(x+100)
=(x+100-x)/[x(x+100)]
=100/(x^2+100x).

[1/x(x+1)]+[1/(x+1)(x+2)]+[1/(x+2)(x+3)]+... ...+[1/(x+99)(x+100)]
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+…+1/(x+99)-1/(x+100)
=1/x-1/(x+100)
=100/(x^2+100x)

原式=1/x-1/(x+1)+1/(x+1)-1/(x+2).........1/(x+99)- 1/(x+100)
=1/x-1/(x+100)
=100/x(x+100)

=1/x-1/(x+1)+1/(x+1)-1/(x+2).........1/(x+99)-1/(x+100)
=1/x-1/(x+100)

每一项都可以分解成两下相减。如1/x(x+1)=1/x -1/(x+1）
则原式=1/x-1/(x+1)+1/(x+1)-1/(x+2).........1/(x+99)-1/(x+100)
=1/x-1/(x+100)

这种题都是这个规律，先把每项分解，然后前后抵消，就会得出答案了
祝学习进步