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