整理（a-3）/(3a²-6a)÷[a+2-5/(a-2)]
=（a-3）/3a（a-2)÷[(a+2)(a-2)/a-2-5/(a-2)]
=（a-3）/3a（a-2)÷[(a+2)(a-2)-5/(a-2)]
=（a-3）/3a（a-2)÷[(a²-9)/(a-2)]
=（a-3）/3a（a-2)*(a-2)/(a+3)(a-3)
=1/3a(a+3)
=1/3(a²+3a)
因为a是一元二次方程X^2+3X-1=0的实数根
所以a²+3a-1=0,
所以a²+3a=1,代人,得,
原式=1/3