令a=2000
则根号下=(a-1)a(a+1)(a+2)+1
=[(a-1)(a+2)][a(a+1)]+1
=[(a²+a)-2](a²+a)+1
=(a²+a)²-2(a²+a)+1
=(a²+a-1)²
所以原式=a²+a-1-a²
=a-1
=1999
令a=2000
则根号下=(a-1)a(a+1)(a+2)+1
=[(a-1)(a+2)][a(a+1)]+1
=[(a²+a)-2](a²+a)+1
=(a²+a)²-2(a²+a)+1
=(a²+a-1)²
所以原式=a²+a-1-a²
=a-1
=1999