回复(2) 2010-05-20 10:35 来自版块 - 
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beike多谢T2 (2010-05-26 11:21)
teacher2解答:-ax^2+3x+2a-1>0对x∈(0,1]恒成立 (2-x^2)a+3x-1>0 a>(1-3x)/(2-x^2) 设1-3x=t∈[-2,1) (1-3x)/(2-x^2)=t/[2-(1-t)^2/9] =1/(17/9t-t/9+2/9...(2010-05-23 01:28)

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