怎么解?(1)at^2+8t+8=0,a-8求t最大值
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 17:51:39
![怎么解?(1)at^2+8t+8=0,a-8求t最大值](/uploads/image/z/14755782-30-2.jpg?t=%E6%80%8E%E4%B9%88%E8%A7%A3%3F%EF%BC%881%EF%BC%89at%5E2%2B8t%2B8%3D0%2Ca-8%E6%B1%82t%E6%9C%80%E5%A4%A7%E5%80%BC)
怎么解?(1)at^2+8t+8=0,a-8求t最大值
怎么解?(1)at^2+8t+8=0,a<= -8 求t的最大值(2)at^2+8t-2=0,a>-8求t最大值
怎么解?(1)at^2+8t+8=0,a-8求t最大值
(1)由at^2+8t+8=0,可知t不等于0,则a=-(8t+8)/t^2《-8,
解得t的最大值是(1+根号5)/2
(2)由at^2+8t-2=0,可知t不等于0,则a=-(8t-2)/t^2>-8,
解得t不等于1/2,所以此时t没有最大值