一道代数题，麻烦要快！~~

因为m、n互为倒数
所以m+1/m+2000=0
m^2+2000m+1=0
1=-m^2-2000m
同理1=-n^2-2000n
所以(m^2+2001m+1)(n^2+2001n+1)
=(m^2+2001m-m^2-2000m)(n^2+2001n-n^2-2000n)
=mn
=1

m、n互为倒数,mn=1
m+n+2000=0----m^2+n^2+2000^2=0
(m^2+2001m+1)(n^2+2001n+1)
=m^2n^2+2001m^2n+m^2+2001mn^2+(2001)^2mn+2001m+n^2+2001n+1
=1+2001m+m^2+2001n+(2001)^2+2001m+n^2+2001n+1
=2+m^2+n^2+2001(2m+2n+2001)
=2+m^2+n^2+2001[2(m+n+2000)+1999]
=2+m^2+n^2+2001*1999
=2+m^2+n^2+(2000+1)(2000-1)
=2+m^2+n^2+2000^2-1
=2-1=1