已知a是一元二次方程x²+3x-1=0的实数根,求代数式（x-3/3x²-6x）÷[x+2-（5/x

x²+3x=1
所以原式=(x-3)/3x(x-2)×(x-2)/(x²-4-5)
=(x-3)/3x×1/(x+3)(x-3)
=1/3x(x+3)
=1/[3(x²+3x)]
=1/3