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y=(4t+3)/(t²+1)值域怎么求
人气:269 ℃ 时间:2020-06-17 11:03:46
解答
用判别式算:
y=(4t+3)/(t^2+1)
(t^2+1)y=4t+3
y•t^2-4t+(y-3)=0
①当y=0时 t= -3/4
②当y≠0时 △>=0
(-4)^2-4y(y-3)>=0
-4y^2+12y+16>=0
(y-4)(y+1)
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