1...1/n(n+1)=1/n-1/(n+1)
证明:1/n-1/(n+1)=[(n+1)-n]/[n(n+1)]=1/n(n+1)
2...1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2)
=1/(x-3)-1/(x-2)-2/(x-1)(x-3)+1/(x-2)-1/(x-1)
=1/(x-3)-2/(x-1)(x-3)-1/(x-1)
=0
1...1/n(n+1)=1/n-1/(n+1)
证明:1/n-1/(n+1)=[(n+1)-n]/[n(n+1)]=1/n(n+1)
2...1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2)
=1/(x-3)-1/(x-2)-2/(x-1)(x-3)+1/(x-2)-1/(x-1)
=1/(x-3)-2/(x-1)(x-3)-1/(x-1)
=0