Question

10．已知$\left\{\begin{array}{l}{{k}^{2}-2≠0}\\{-\frac{2k}{{k}^{2}-2}＞0}\\{\frac{2}{{k}^{2}-2}＞0}\end{array}\right.$，求k的取值范围．

Answer 1

分析 原不等式等价为$\left\{\begin{array}{l}{{k}^{2}-2＞0}\\{k＜0}\end{array}\right.$，由二次不等式的解法即可得到所求范围．

解答 解：原不等式等价为$\left\{\begin{array}{l}{{k}^{2}-2＞0}\\{k＜0}\end{array}\right.$，
即为$\left\{\begin{array}{l}{k＞\sqrt{2}或k＜-\sqrt{2}}\\{k＜0}\end{array}\right.$，
解得k＜-$\sqrt{2}$．
则k的取值范围是（-∞，-$\sqrt{2}$）．

点评 本题考查分式不等式的解法，注意转化为整式不等式，考查运算能力，属于基础题．