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若一元二次不等式a 对xR恒成立,求求数a的-|||-ax^2-
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答案a<-1/3,a的范围是小于负3分之1,解析:二次函数小于零,x属于R,恒成立,那么a0,求出a<-1/3
咨询记录 · 回答于2023-02-15
若一元二次不等式a 对xR恒成立,求求数a的-|||-ax^2-
你好,同学,麻烦把原题的截图发出来,这样能够快速有效地解决问题![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
答案a<-1/3,a的范围是小于负3分之1,解析:二次函数小于零,x属于R,恒成立,那么a0,求出a<-1/3
答案a<-1/3,a的范围是小于负3分之1,解析:二次函数小于零,x属于R,恒成立,那么a0,求出a<-1/3
解题方法:首先画图,二次函数在x轴下方,开口向下,然后与x轴没有交点,判别式小于零
详细过程请看截图![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
好了吗
答案发了2遍![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
没看见
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_11.png)
看见吧
眼睛大
好的
详细过程请看截图![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)
![](https://s.bdstatic.com/common/openjs/emoticon/img/face_01.png)