y=(2x·x-x+2)/(x·x+x+1),用判别式法求函数值域
人气：418 ℃ 时间：2019-10-20 20:15:47
解答
yx^2+yx+y=2x^-x+2
(y-2)x^2+(y+1)x+(y-2)=0
△>=0
所以y^2+2y+1-4y^2+16>=0
3y^2-2y-17
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