Question

3．關(guān)于x的不等式1og${\;}_{\frac{1}{2}}$（x²-8）＞1og${\;}_{\frac{1}{2}}$2x的解集為（$2\sqrt{2}，4$）．

Answer 1

分析 由對(duì)數(shù)函數(shù)的單調(diào)性化對(duì)數(shù)不等式為一元二次不等式組求解．

解答 解：由1og${\;}_{\frac{1}{2}}$（x²-8）＞1og${\;}_{\frac{1}{2}}$2x，得$\left\{\begin{array}{l}{{x}^{2}-8＞0}\\{2x＞{x}^{2}-8}\end{array}\right.$，解得$2\sqrt{2}＜x＜4$．
∴不等式1og${\;}_{\frac{1}{2}}$（x²-8）＞1og${\;}_{\frac{1}{2}}$2x的解集為（$2\sqrt{2}，4$）．
故答案為：（$2\sqrt{2}，4$）．

點(diǎn)評(píng) 本題考查對(duì)數(shù)不等式的解法，考查對(duì)數(shù)函數(shù)的單調(diào)性，是基礎(chǔ)題．