Задача 12А. Решите неравенство \(\left| {\frac{{x-1}}{{x + 2}}} \right| > 1\)
ОТВЕТ: \(\left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-\frac{1}{2}} \right).\)
\(\left| {\frac{{x-1}}{{x + 2}}} \right| > 1.\) Неравенство вида \(\left| {f\left( x \right)} \right| > a\) при \(a > 0\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) > a,\,\,}\\{f\left( x \right) < -a.}\end{array}} \right.\) \(\left| {\frac{{x-1}}{{x + 2}}} \right| > 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\frac{{x-1}}{{x + 2}} > 1,\,}\\{\frac{{x-1}}{{x + 2}} < -1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\frac{{x-1}}{{x + 2}}-1 > 0,\,}\\{\frac{{x-1}}{{x + 2}} + 1 < 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\frac{{-3}}{{x + 2}} > 0,\,}\\{\frac{{2x + 1}}{{x + 2}} < 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x \in \left( {-\infty ;\,-2} \right),}\\{x \in \left( {-2;\,-\frac{1}{2}} \right)}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-\frac{1}{2}} \right).\) Ответ: \(\left( {-\infty ;\;-2} \right) \cup \left( {-2;\;-\frac{1}{2}} \right).\)