\(\left| {2x-1} \right| \le \left| {4x + 1} \right|.\)
Неравенство вида \(\left| {f\left( x \right)} \right| \le \left| {g\left( x \right)} \right|\) решается возведением обеих частей неравенства в квадрат (это допустимо, так как обе части неравенства неотрицательны):
\({f^2}\left( x \right) \le {g^2}\left( x \right)\;\;\;\; \Leftrightarrow \;\;\;\;{f^2}\left( x \right)-{g^2}\left( x \right) \le 0\;\;\;\; \Leftrightarrow \;\;\;\;\left( {f\left( x \right)-g\left( x \right)} \right)\left( {f\left( x \right) + g\left( x \right)} \right) \le 0.\)
\(\left| {2x-1} \right| \le \left| {4x + 1} \right|\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {2x-1} \right)^2} \le {\left( {4x + 1} \right)^2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {2x-1} \right)^2}-{\left( {4x + 1} \right)^2} \le 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {2x-1-4x-1} \right)\left( {2x-1 + 4x + 1} \right) \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {-2x-2} \right) \cdot 6x \le 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\left( {x + 1} \right) \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\)
\( \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;-1} \right] \cup \left[ {0; + \infty } \right)\)
Ответ: \(\left( {-\infty ;-1} \right] \cup \left[ {0; + \infty } \right).\)