问题
填空题
设对所有实数x，不等式x2log2
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答案
∵不等式x2log2
+2xlog24(a+1) a
+log22a a+1
＞0恒成立(a+1)2 4a2
由二次不等式的性质可得，log2
＞0且△=4(log24(a+1) a
)2-log22a a+1
•log24(a+1) a
×4＜0(a+1)2 4a2
令t=log2a+1 a
即(1+log2
)2-(2+log2a a+1
)(2log2a+1 a
-2)＜0a+1 a
整理可得，(log2
+5)(log2a+1 a
-1)＞0a+1 a
∵log2
＞04(a+1) a
∴log2
＞1a+1 a
解可得，0＜a＜1
故答案为：0＜a＜1