Задача 43. Решите уравнение \(\left| {\,4x-\left| {x-2} \right| + 3\,} \right| = 16\)
ОТВЕТ: -17/5; 11/3.
\(\left| {4x-\left| {x-2} \right| + 3} \right| = 16.\) Уравнение вида \(\left| {f\left( x \right)} \right| = a,\) где \(a \ge 0,\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = a,\;\,}\\{f\left( x \right) = -a.}\end{array}} \right.\) \(\left| {4x-\left| {x-2} \right| + 3} \right| = 16\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{4x-\left| {x-2} \right| + 3 = 16,\;\,}\\{4x-\left| {x-2} \right| + 3 = -16}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left| {x-2} \right| = 4x-13,}\\{\left| {x-2} \right| = 4x + 19}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\) \( \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{-x + 2 = 4x-13,}\\{-x + 2 = 4x + 19}\end{array}} \right.}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \ge 2,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x-2 = 4x-13,}\\{x-2 = 4x + 19}\end{array}} \right.}\end{array}\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < 2,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = 5,}\\{x = -\dfrac{{17}}{5}}\end{array}} \right.}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \ge 2,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{11}}{3},}\\{x = -7}\end{array}} \right.}\end{array}\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{{17}}{5},}\\{x = \dfrac{{11}}{3}.\,\,\,\,}\end{array}} \right.\) Ответ: \(-\dfrac{{17}}{5};\;\;\,\;\dfrac{{11}}{3}.\)