已知A是一元二次方程x²+3X-1=0的根,那么代数式A-3／3A²-6A÷（A+2-5／A-2）=

x=a
a²+3a-1=0
a²+3a=1
原式=(a-3)/3a(a-2)÷[(a²-4-5)/(a-2)]
=(a-3)/3a(a-2)×(a-2)(a+3)(a-3)
=1/3a(a+3)
=1/[3(a²+3a)]
=1/3