Профиль №15. Неравенства с модулями. Задача 18А

ОТВЕТ: \(\left( {-\infty ;\;2} \right] \cup \left[ {4;\;\infty } \right).\)

\(\left| {3x-7} \right| \ge 2x-3.\)

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

\(\left| {3x-7} \right| \ge 2x-3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{3x-7 \le -2x + 3,}\\{3x-7 \ge 2x-3\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x \le 2,}\\{x \ge 4}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;2} \right] \cup \left[ {4; + \infty } \right).\)

Ответ:  \(\left( {-\infty ;2} \right] \cup \left[ {4; + \infty } \right)\).