(ax+1)/(x^2+4x+3)<0的解为x<-4或-3<x<-1,求a的取值范围.

热心网友

(ax+1)/(x^2+4x+3)(ax+2)/[(x+1)(x+3)](ax+1)(x+1)(x+3)a=1/4

热心网友

化为(ax+1)(x+1)(x+3)<0穿根法-4a+1=0a=0.25