问题补充:
填空题若命题“任意的x∈R,x2+ax+1≥0”是假命题,则实数a的取值范围是________.
答案:
(-∞,-2)∪(2,+∞)解析分析:由命题“任意的x∈R,x2+ax+1≥0”是假命题,知△=a2-4>0,由此能求出实数a的取值范围.解答:∵命题“任意的x∈R,x2+ax+1≥0”是假命题,∴△=a2-4>0,∴a>2或a<-2.故
时间:2021-12-31 21:13:27
填空题若命题“任意的x∈R,x2+ax+1≥0”是假命题,则实数a的取值范围是________.
(-∞,-2)∪(2,+∞)解析分析:由命题“任意的x∈R,x2+ax+1≥0”是假命题,知△=a2-4>0,由此能求出实数a的取值范围.解答:∵命题“任意的x∈R,x2+ax+1≥0”是假命题,∴△=a2-4>0,∴a>2或a<-2.故
填空题命题p:?x0∈R +ax0+1≤0为假命题 则实数a的取值范围是_______
2019-06-12