题目

设对所有实数x，不等式x2log2

4(a+1)

a

+2xlog2

2a

a+1

+log2

(a+1)2

4a2

＞0恒成立，则a的取值范围为______．

答案

∵不等式x2log2

4(a+1)

a

+2xlog2

2a

a+1

+log2

(a+1)2

4a2

＞0恒成立
由二次不等式的性质可得，log2

4(a+1)

a

＞0且△=4(log2

2a

a+1

)2-log2

4(a+1)

a

•log2

(a+1)2

4a2

×4＜0
令t=log₂

a+1

a

即(1+log2

a

a+1

)2-(2+log2

a+1

a

)(2log2

a+1

a

-2)＜0
整理可得，(log2

a+1

a

+5)(log2

a+1

a

-1)＞0
∵log2

4(a+1)

a

＞0
∴log2

a+1

a

＞1
解可得，0＜a＜1
故答案为：0＜a＜1

解析