Задача 22А. Решите неравенство \(\left| {\left| {2x-1} \right|-2} \right| > 3\)
ОТВЕТ: \(\left( {-\infty ;\;-2} \right) \cup \left( {3;\;\infty } \right).\)
\(\left| {\left| {2x-1} \right|-2} \right| > 3.\) Неравенство вида \(\left| {f\left( x \right)} \right| > a\) при \(a > 0\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) > a,\,\,}\\{f\left( x \right) < -a.}\end{array}} \right.\) \(\left| {\left| {2x-1} \right|-2} \right| > 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {2x-1} \right|-2 < -3,}\\{\left| {2x-1} \right|-2 > 3\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {2x-1} \right| < -1,}\\{\left| {2x-1} \right| > 5\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left| {2x-1} \right| > 5\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x-1 < -5,}\\{2x-1 > 5\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x < -4,}\\{2x > 6\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x < -2,}\\{x > 3\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;-2} \right) \cup \left( {3; + \infty } \right).\) Ответ: \(\left( {-\infty ;-2} \right) \cup \left( {3; + \infty } \right)\).