Профиль №15. Неравенства с модулями. Задача 19А

Задача 19А. Решите неравенство \(1 \leqslant \left| {x-2} \right| < 3\)

Ответ

ОТВЕТ: \(\left( {-1;\;1} \right] \cup \left[ {3;\;5} \right)\)

Решение

\(1 \le \left| {x-2} \right| < 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left| {x-2} \right| < 3,}\\{\left| {x-2} \right| \ge 1\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-3 < x-2 < 3,}\\{\left[ {\begin{array}{*{20}{c}}{x-2 \ge 1,\,\,}\\{x-2 \le -1}\end{array}\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-1 < x < 5,}\\{\left[ {\begin{array}{*{20}{c}}{x \ge 3,\,\,}\\{x \le 1}\end{array}\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-1 < x < 5,\,\,\,\,\,\,\,\,\,\,}\\{\,\left( {-\infty ;1} \right] \cup \left[ {3;\,\infty } \right)}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \,\left( {-1;\,1} \right] \cup \left[ {3;\,5} \right).\)

Ответ: \(\left( {-1;1} \right] \cup \left[ {3;5} \right)\).