Профиль №15. Неравенства с модулями. Задача 17А

ОТВЕТ: \(\left( {-\infty ;\;1} \right) \cup \left( {4;\;\infty } \right).\)

\(\left| {3x-6} \right| > x + 2.\)

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

\(\left| {3x-6} \right| > x + 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x-6 < -x-2,}\\{3x-6 > x + 2\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x < 1,}\\{x > 4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;1} \right) \cup \left( {4; + \infty } \right).\)

Ответ:  \(\left( {-\infty ;1} \right) \cup \left( {4; + \infty } \right)\).