Задача 6С. Решите уравнение \(\left| {\,{x^2}-\left| x \right|-6\,} \right| = \left| x \right| + 2\)
ОТВЕТ: \( \pm 4;\; \pm 2.\)
\(\left| {\,{x^2}-\left| x \right|-6\,} \right| = \left| x \right| + 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left| {{{\left| x \right|}^2}-\left| x \right|-6} \right| = \left| x \right| + 2.\) Пусть \(\left| x \right| = t\), где \(t \ge 0\). Тогда уравнение примет вид: \(\left| {{t^2}-t-6} \right| = t + 2.\) Так как \(t \ge 0\), то \(t + 2 > 0\). Поэтому последнее уравнение равносильно совокупности. \(\left[ {\begin{array}{*{20}{c}}{{t^2}-t-6 = t + 2\,\,}\\{{t^2}-t-6 = -t-2}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{t^2}-2t-8 = 0}\\{{t^2} = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -2 < 0,}\\{t = 4,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = 2.\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\) Вернёмся к переменной x: \(\left[ {\begin{array}{*{20}{c}}{\left| x \right| = 4,}\\{\left| x \right| = 2\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm 4,}\\{x = \pm 2.}\end{array}} \right.\) Ответ: \( \pm 4;\; \pm 2.\)