已知a是一元二次方程x2+3x-2=0的实数根,求代数式a-3/3a2-6a÷(a+2-5/a-2)的值

Question

如题所述

Answer 1

a是一元二次方程x2+3x-2=0的实数根
∴a²+3a=2
(a-3)/(3a2-6a)÷[(a+2)-5/(a-2)]
=(a-3)/[3a(a-2)]÷[(a²-4-5)/(a-2)]
=(a-3)/[3a(a-2)]×(a-2)/[(a+3)(a-3)]
=1/[(a+3)(3a)]
=1/[3(a²+3a)]
=1/(3*2)
=1/6