(a2-4/a2-4a+4-1/2-a)/2/a2-2a,其中a是方程x2+3x+1的根
问题描述：
(a2-4/a2-4a+4-1/2-a)/2/a2-2a,其中a是方程x2+3x+1的根
a2-4/a2-4a+4 1/2-a2/a2-2a分别是1组啊，别看错了
答
即a²+3a+1=0
a²+3a=-1
原式=[(a+2)(a-2)/(a-2)²+1/(a-2)]/[2/a(a-2)]
=[(a+2)/(a-2)+1/(a-2)]×[a(a-2)/2]
=[(a+3)/(a-2)]×[a(a-2)/2]
=a(a+3)/2
=(a²+3a)/2
=-1/2