已知一元二次方程x²＋3x－1＝0的一个实数根,求x-3/3x²-6x÷（x+2-5/x-2）的值

(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+3)(x-3)/(x-2)]=(x-3)/[3x(x-2)]×(x-2)/[(x+3)(x-3)]=1/[3x(x+3)]=1/[3(x²+3x)]∵x²+3x-1=0∴x...