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得：1/1*2+1/2*3+1/3*4=1-1/2+1/2-1/3+1/3-1/4=1-1/4=3/4.（1）直接写出下列各式的计算结果:1/1*2+1/2*3+1/3*4+.+1/n(n+1)= (2)猜想并写出；1/n(n+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

这是分式相消的思想!
观察结果：
注意：
1/1*2 = (2-1)/1*2 = 2/1*2 - 1/1*2 = 1/1 - 1/2
1/2*3 = (3-2)/2*3 = 3/2*3 - 2/2*3 = 1/2 - 1/3
1/3*4 = (4-3)/3*4 = 4/3*4 - 3/3*4 = 1/3 - 1/4
上述各式相加：
左边=1/1*2+1/2*3+1/3*4
右边=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 = 1- 1/4 = 3/4
因此：
1）
:1/1*2+1/2*3+1/3*4+.+1/n(n+1)=1 - 1/(n+1)
2)
1/n(n+2)= = (1/2)* [(n+2)-n]/n(n+2) = (1/2)* [1/n - 1/(n+2)]
3)
1/x(x+3) = (1/3)*[(x+3)-x]/x(x+3)=(1/3)*[1/x - 1(x+3)]
1/(x+3)(x+6) =(1/3)*[(x+6)-(x+3)]/(x+3)(x+6)=(1/3)*[1/(x+3) - 1(x+6)]
1/(x+6)(x+9) =(1/3)*[(x+3)-(x+6)]/(x+6)(x+9)=(1/3)*[1/(x+6) - 1(x+9)]
所以：
1/x(x+3)+1/(x+3)(x+6)+1/(x+6)(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)]
=(1/3)*[(x+9-x)/x(x+9)]
=3/x(x+9)
=3/(2x+18) (你写的不清楚,就这么理解了!)
=3/2(x+9)
原式化简成：
3/x(x+9) = 3/2(x+9)
1/x = 1/2
所以：
x=2