原式=a/(a+1)-(a+3)/(a-1)*(a-1)²/(a+1)(a+3)
=a/(a+1)-(a-1)/(a+1)
=(a-a+1)/(a+1)
=1/(a+1)
∵a是方程a^2+3a-1=0的一个根
∴a=(-3±√13)/2
∴a=(-3-√13)/2时,原式=1/[(-3-√13)/2+1]=1/[(-1-√13)/2]=2/(-1-√13)=(1-√13)/6
a=(-3+√13)/2时,原式=1/[(-3+√13)/2+1]=1/[(-1+√13)/2]=2/(-1+√13)=(1+√13)/6