已知x是一元二次方程x²+3x-1=0的实数根
则x²+3x=1
(x-3)/(3x²-6x)÷[x+2- 5/(x-2)]
=(x-3)/[3x(x-2)]÷[(x²-4-5)/(x-2)]
=[(x-3)/3x(x-2)]*(x-2)/[(x+3)(x-3)]
=1/[3x(x+3)]
=1/[3(x²+3x)
=1/3