Задача 3С. Решите уравнение \(\left| {\,\left| {\,\left| {\,\left| {\,x-1\,} \right| + 2\,} \right|-1\,} \right| + 1\,} \right| = 2\)
ОТВЕТ: 1.
\(\left| {\,\left| {\,\left| {\,\left| {\,x-1\,} \right| + 2\,} \right|-1\,} \right| + 1\,} \right| = 2.\) Уравнение вида \(\left| {f\left( x \right)} \right| = a\), где \(a \ge 0\), равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = a,\,\,\,}\\{f\left( x \right) = -a.}\end{array}} \right.\) \(\left| {\left| {\left| {\left| {x-1} \right| + 2} \right|-1} \right| + 1} \right| = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {\left| {\left| {x-1} \right| + 2} \right|-1} \right| + 1 = 2,\,\,}\\{\left| {\left| {\left| {x-1} \right| + 2} \right|-1} \right| + 1 = -2}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {\left| {\left| {x-1} \right| + 2} \right|-1} \right| = 1,\,\,\,}\\{\left| {\left| {\left| {x-1} \right| + 2} \right|-1} \right| = -3}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {\left| {x-1} \right| + 2} \right|-1 = 1,\,\,\,}\\{\left| {\left| {x-1} \right| + 2} \right|-1 = -1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {\left| {x-1} \right| + 2} \right| = 2,}\\{\left| {\left| {x-1} \right| + 2} \right| = 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {x-1} \right| + 2 = 2,\,\,\,}\\{\left| {x-1} \right| + 2 = -2,}\\{\left| {x-1} \right| + 2 = 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {x-1} \right| = 0,\,\,\,\,}\\{\left| {x-1} \right| = -4,}\\{\left| {x-1} \right| = -2\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,x-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 1.\) Ответ: 1.