令x=2005
则根号内=a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)+1
=(a²+3a)(a²+3a+2)+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)²
所以原式=a²+3a+1-a²
=3a+1
=6016
令x=2005
则根号内=a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)+1
=(a²+3a)(a²+3a+2)+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)²
所以原式=a²+3a+1-a²
=3a+1
=6016