Question

3．不等式$\frac{x-1}{{x}^{2}-1}$＞0的解是{x|x＞1或-1＜x＜1}．

Answer 1

分析 不等式$\frac{x-1}{{x}^{2}-1}$＞0等價于$\left\{\begin{array}{l}{{x}^{2}-1≠0}\\{x-1＞0}\\{{x}^{2}-1＞0}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{2}-1≠0}\\{x-1＜0}\\{{x}^{2}-1＜0}\end{array}\right.$，解不等式組可得．

解答 解：不等式$\frac{x-1}{{x}^{2}-1}$＞0等價于$\left\{\begin{array}{l}{{x}^{2}-1≠0}\\{x-1＞0}\\{{x}^{2}-1＞0}\end{array}\right.$或$\left\{\begin{array}{l}{{x}^{2}-1≠0}\\{x-1＜0}\\{{x}^{2}-1＜0}\end{array}\right.$，
解不等式組可得x＞1或-1＜x＜1，
∴原不等式的解集為{x|x＞1或-1＜x＜1}，
故答案為：{x|x＞1或-1＜x＜1}．

點評 本題考查分式不等式的解集，轉(zhuǎn)化為不等式組是解決問題的關(guān)鍵，屬基礎(chǔ)題．