观察下列各式，1⼀6=1⼀2*3=1⼀2-1⼀3;1⼀12=1⼀3*4=1⼀3-1⼀4:1⼀20=1⼀4*5=1⼀4-1⼀5;1⼀30=1⼀5*6=1⼀5-1⼀6......

1/(n*(n+1))=1/n-1/(n+1)
证明：1/(n*(n+1))=1/n-1/(n+1)=（n+1-n)/(n*(n+1))=1/(n*(n+1))

这方程怎么没有等号？？？？
1/（x-2)（x-3)-2/(x-1)(x-3)+1/(x-1)(x-2)=1/（x-2)（x-3)-1/(x-1)(x-3)+)-1/(x-1)(x-3)+1/(x-1)(x-2)
=1/（x-3)*(1/（x-2)-1/(x-1))+1/(x-1)*(1/(x-2)-1/(x-3))
=-1/（（x-3)*（（x-2)(x-1))）+1/（（x-3)*（（x-2)(x-1))）=0

解：
公式：
1/n(n+1) = 1/n - 1/(n+1)，其中：n为正整数

证明：
1/n(n+1)
=[(n+1) - n] / n(n+1)
=(n+1) / n(n+1) - n / n(n+1)
=1/n - 1/(n+1)

解：
(2)
1/(x-2)(x-3) = [(x-2)-(x-3)] / (x-2)(x-3) = (x-2)/(x-2)(x-3) - (x-3)/(x-2)(x-3)
=1(x-3) - 1/(x-2)
-2/(x-1)(x-3) =-[(x-1)-(x-3)] / (x-1)(x-3) = (x-2) (x-2)(x-3) - (x-3) (x-1)(x-3)
= -1/(x-3) + 1/(x-1)
1/(x-1)(x-2) =[(x-1)-(x-2)]/(x-1)(x-2) = (x-1)/(x-1)(x-2) - (x-2)/(x-1)(x-2)
=1/(x-2) - 1/(x-1)
上述各式相加：
左边=1/(x-3) - 1/(x-2) - 1/(x-3) + 1/(x-1) + 1/(x-2) - 1/(x-1)
= 0