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[log2(x)]^2+(a+1/a)log1/2(x)+1<0
人气:333 ℃ 时间:2020-05-09 00:44:16
解答
log2(x)]^2-(a+1/a)log2(x)+1<0.令log2(x)=m,则m^2-(a+1/a)m+1<0.△=(a-1/a)^2.m^2-(a+1/a)m+1=0m为a或1/a.m^2-(a+1/a)当m+1<0a>1或a<-1.1/a^2
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