Question

19．解不等式$\frac{{x}^{2}-2x-1}{x-1}$≥0．

Answer 1

分析 原不等式等價于$\left\{\begin{array}{l}{x-1＞0}\\{{x}^{2}-2x-1≥0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜0}\\{{x}^{2}-2x-1≤0}\end{array}\right.$，分別解不等式組取并集可得．

解答 解：原不等式等價于$\left\{\begin{array}{l}{x-1＞0}\\{{x}^{2}-2x-1≥0}\end{array}\right.$或$\left\{\begin{array}{l}{x-1＜0}\\{{x}^{2}-2x-1≤0}\end{array}\right.$，
解不等式組$\left\{\begin{array}{l}{x-1＞0}\\{{x}^{2}-2x-1≥0}\end{array}\right.$可得{x|x≥1+$\sqrt{2}$}
解不等式組$\left\{\begin{array}{l}{x-1＜0}\\{{x}^{2}-2x-1≤0}\end{array}\right.$可得{x|x≤1-$\sqrt{2}$}，
綜合可得原不等式的解集為{x|x≥1+$\sqrt{2}$或x≤1-$\sqrt{2}$}

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，屬中檔題．