(1)
令a²+4a-5=0
(a+5)(a-1)=0
a=-5或a=1
a=-5时,不等式变为8x<1x<1/8,与已知不符,a=-5不满足题意.
a=1时,不等式变为3>0,不等式恒成立,x可取任意实数,a=1满足题意.
(2)
a≠1且a≠-5时,要不等式恒成立,则二次项系数a²+4a-5>0,方程(a²+4a-5)x²+4(a-1)x+3=0判别式<0
a²+4a-5>0(a+5)(a-1)>0a>1或a<-5
[4(a-1)]²-12(a²+4a-5)<0
a²-20a+19<0
(a-1)(a-19)<0
1
综上,得1≤a<19