Неравенства с модулями. Задача 7

ОТВЕТ: \(\left( {-\infty ;\;-6} \right) \cup \left( {2;\;\infty } \right).\)

\(\left| {x + 2} \right| > 4.\)

Неравенство вида  \(\left| {f\left( x \right)} \right| > a\) при \(a > 0\)  равносильно совокупности:  \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) > a,\,\,}\\{f\left( x \right) < -a.}\end{array}} \right.\)

\(\left| {x + 2} \right| > 4\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 2 > 4,\,\,}\\{x + 2 < -4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x > 2,\,}\\{x < -6}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\;-6} \right) \cup \left( {2;\; + \infty } \right).\)

Ответ:  \(\left( {-\infty ;\;-6} \right) \cup \left( {2;\; + \infty } \right).\)