\(\left| {{x^3}-4x} \right| + \left| {4{x^2}-1} \right| = {x^3} + 4{x^2}-4x-1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left| {{x^3}-4x} \right| + \left| {4{x^2}-1} \right| = {x^3}-4x + 4{x^2}-1.\)
Пусть \({x^3}-4x = a,\,\,\,\,\,4{x^2}-1 = b.\) Тогда уравнение примет вид: \(\left| a \right| + \left| b \right| = a + b.\)
Решение последнего уравнения: \(\left\{ {\begin{array}{*{20}{c}}{a \ge 0,}\\{b \ge 0.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left\{ {\begin{array}{*{20}{c}}{{x^3}-4x \ge 0}\\{4{x^2}-1 \ge 0\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\left( {x-2} \right)\left( {x + 2} \right) \ge 0}\\{\left( {2x-1} \right)\left( {2x + 1} \right) \ge 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x\, \in \,\left[ {-2;0} \right] \cup \left[ {2;\infty } \right),\,\,\,\,\,\,\,\,\,}\\{x\, \in \,\left( {-\infty ;-\dfrac{1}{2}} \right] \cup \left[ {\dfrac{1}{2};\infty } \right)}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left[ {-2;-\dfrac{1}{2}} \right] \cup \left[ {2;\infty } \right).\)
Ответ: \(\left[ {-2;\;-\dfrac{1}{2}} \right] \cup \left[ {2;\;\infty } \right).\)