已知关于x的不等式(a^2+4a-5)x^2-4(a-1)x+3>0的解集为R,则实数a的取值范围是___
咋做?
人气:204 ℃ 时间:2020-02-05 19:14:02
解答
(a^2+4a-5)x^2-4(a-1)x+3>0
(1)当a=1时
0*x^2-0*x+3=3>0
解集为R
(2)当a^2+4a-5>0
(a+5)(a-1)>0
a1
且Δ=16(a-1)^2-12(a^2+4a-5)
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