当x≥1时,|x+3|-|x-1|=x+3-(x-1)=4;
当-3≤x<1时,|x+3|-|x-1|=x+3-(1-x)=2x+2<4;
当x<3时,|x+3|-|x-1|=-x-3-(1-x)=-4;
所以,|x+3|-|x-1|≤4恒成立;
据题意可得a^2-3a>4;
所以a>4或a<-1.
不等式Ix+3I-Ix-1I<a²-3a对任意实数x恒成立,则实数a的取值范围
不等式Ix+3I-Ix-1I<a²-3a对任意实数x恒成立,则实数a的取值范围
数学人气:244 ℃时间:2020-02-03 22:32:23
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