题目

(1)1/2=1/(1*2)=1/1-1/2
1/6=1/(2*3)=1/2-1/3
1/12=1/(3*4)=1/3-1/4
1/20=1/(4*5)=1/4-1/5
.
用含字母n的等式表示出上面式子的规律,并证明.
并用写出的规律计算:
1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2).
(2)已知(x^2-1)/(x-2)(x-3)=a+b/(x-2)+c/(x-3),求a+b+c的值.

提问时间：2021-02-16

答案

(1)
1/[n*(n+1)]=1/n-1/(n+1)
证明：
1/n-1/(n+1)
=(n+1)/[n*(n+1)]-n/[n*(n+1)]
=1/[n*(n+1)]
1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2).
=[1/(x-3)-1/(x-2)]-[1/(x-3)-1/(x-1)]+[1/(x-2)-1/(x-1)]
=1/(x-3)-1/(x-2)-1/(x-3)+1/(x-1)+1/(x-2)-1/(x-1)
=0
(2)
(x^2-1)/(x-2)(x-3)=a+b/(x-2)+c/(x-3)
(x+1)(x-1)/(x-2)(x-3)=a+b/(x-2)+c/(x-3)
[(x+1)/(x-2)]*[(x-1)/(x-3)]=a+b/(x-2)+c/(x-3)
[1+3/(x-2)]*[1+2/(x-3)]=a+b/(x-2)+c/(x-3)
1+3/(x-2)+2/(x-3)+6/(x-2)(x-3)=a+b/(x-2)+c/(x-3)
1+3/(x-2)+2/(x-3)+6/(x-3)-6/(x-2)=a+b/(x-2)+c/(x-3)
1-3/(x-2)+8/(x-3)=a+b/(x-2)+c/(x-3)
a=1,b=-3,c=8