题目

观察下列各式：1/1*2=1/1-1/2,1/2*3=1/2-1/3,1/3*4=1/3-1/4.
(1）利用上述规律计算1/2+1/6+1/12+...+1/（n-1)n + 1/n(n+1)
(2)利用上述规律解方程 1/(x-4)(x-3)+1/(x-3)(x-2）+1/(x-2)（x-1)+1/(x-1)x+1/x(x+1）=1/x+1

提问时间：2021-03-20

答案

(1）利用上述规律计算:
1/2+1/6+1/12+...+1/（n-1)n + 1/n(n+1)
=1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
(2)利用上述规律解方程:
1/(x-4)(x-3)+1/(x-3)(x-2）+1/(x-2)（x-1)+1/(x-1)x+1/x(x+1）=1/(x+1)
1/(x-4)-1/(x-3)+1/(x-3)-1/(x-2)+1/(x-2)-1/(x-1)+1/(x-1)-1/x+1/x-1/(x+1)=1/(x+1)
1/(x-4)-1/(x+1)=1/(x+1)
[x+1-(x-4)]/[(x-4)(x+1)=1/(x+1)
5/[(x-4)(x+1)]=1/(x+1)
5(x+1)=(x-4)(x+1）
5=x-4
x=9