观察下列等式：1/1*2=1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4,将以上三个等式两边分别相加

（1）1/1*2+1/2*3+1/3*4+。+1/n(n+1)=
=1-1/2 + 1/2-1/3 + 1/3-1/4 + 。+1/n - 1/(n+1)
=1 - 1/(n+1)
=n/(n+1)
(2)猜想并写出；1/n(n+2)= (1/2) [1/n - 1/(n+2)] （就是1/n - 1/(n+2) 整个再除以2）
（3）探究并解方程;1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18
1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)
=(1/3)[1/x -1/(x+3)] + (1/3)[1/(x+3) - 1/(x+6)] + (1/3)[1/(x+6) - 1/(x+9)]
=(1/3) [ 1/x -1/(x+3) + 1/(x+3) - 1/(x+6) + 1/(x+6) - 1/(x+9) ]
=(1/3) [ 1/x - 1/(x+9)
所以
(1/3) [ 1/x - 1/(x+9) ] = 3/2x+18 你这里分母是2x+18,分子是3对吧?
1/x - 1/(x+9) = 9/(2x+18) （同时乘以x(2x+18)）
2x+18 - 2x = 9x
9x = 18
x = 2