(a分之(a的平方+1)-2)除以(a的平方+2a)分之(a+2)(a-1)
=(a^2-2a+1)/a×a(a+2)/[(a+2)(a-1)]
=(a-1)^2/(a-1)
=a-1
∵a^2-4=0
∴a=±2
因此,原式=2-1=1,或原式=-2-1=-3
(a分之(a的平方+1)-2)除以(a的平方+2a)分之(a+2)(a-1)
=(a^2-2a+1)/a×a(a+2)/[(a+2)(a-1)]
=(a-1)^2/(a-1)
=a-1
∵a^2-4=0
∴a=±2
因此,原式=2-1=1,或原式=-2-1=-3