来自刘钟淇的问题
怎么求(t^2+(2根号2)t)/(t^2+1)最值
怎么求(t^2+(2根号2)t)/(t^2+1)最值
1回答
2020-08-1909:22
怎么求(t^2+(2根号2)t)/(t^2+1)最值
怎么求(t^2+(2根号2)t)/(t^2+1)最值
a=(t^2+2√2t)/(t^2+1)
at^2+a=t^2+2√2t
(a-1)t^2-2√2t+a=0
t是实数则方程有解
所以判别式大于等于0
所以8-4a(a-1)>=0
a^2-a-2