∵a²+2a-2012=0
∴a²+2a+2012
(a-2/a²+2a-a-1/a²+4a+4)/a-4/a+2
=[(a-2)/a(a+2) -(a-1)/(a+2)²]×(a+2)/(a-4)
=[(a-2)(a+2)-a(a-1)] /a(a+2)² ×(a+2)/(a-4)
=(a-4) /a(a+2)² ×(a+2)/(a-4)
=1/a(a+2)
=1/(a²+2a)
=1/2012