[(a-1)/a-(a-2)/(a+1)]÷(2a²-a)/(a²+2a+1)
=[(a²-1)-(a²-2a)]/a(a+1)·(a+1)²/a(2a-1)
=(2a-1)/a(a+1)·(a+1)²/a(2a-1)
=(a+1)/a²
∵a是方程x的平方-5x-5=0的根
∴a²-5a-5=0
∴a²=5(a+1)
∴原式=(a+1)/5(a+1)=1/5