\(\left| {\left| x \right|-2} \right| \le 1.\)
Неравенство вида \(\left| {f\left( x \right)} \right| \le a\) при \(a \ge 0\) равносильно двойному неравенству:
\(-a \le f\left( x \right) \le a\;\;\;\; \Leftrightarrow \;\;\;\;\,\left\{ {\begin{array}{*{20}{c}}{f\left( x \right) \le a,\;\,}\\{f\left( x \right) \ge -a.}\end{array}} \right.\)
\(\left| {\left| x \right|-2} \right| \le 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-1 \le \left| x \right|-2 \le 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,1 \le \left| x \right| \le 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left[ {-3;-1} \right] \cup \left[ {1;3} \right].\)
Ответ: \(\left[ {-3;-1} \right] \cup \left[ {1;3} \right]\).