Задача 4С. Решите уравнение \(\left| {\,\left| {\,2x-1\,} \right|-5\,} \right| + x = \left| {6-x} \right|\)
ОТВЕТ: \(\left[ {\dfrac{1}{2};\;3} \right].\)
\(\left| {\,\left| {\,2x-1\,} \right|-5\,} \right| + x = \left| {6-x} \right|.\) Рассмотрим случаи, когда \(x < 6\) и \(x \ge 6:\) \(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {\left| {2x-1} \right|-5} \right| + x = 6-x\,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \ge 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {\left| {2x-1} \right|-5} \right| + x = -6 + x}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {\left| {2x-1} \right|-5} \right| = 6-2x,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \ge 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {\left| {2x-1} \right|-5} \right| = -6\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.}\end{array}} \right.\) Так как \(-6 < 0\), то вторая система последней совокупности не имеет решений. \(\left\{ {\begin{array}{*{20}{c}}{x < 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {\left| {2x-1} \right|-5} \right| = 6-2x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x < 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{6-2x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{\left| {2x-1} \right|-5 = 6-2x,}\\{\left| {2x-1} \right|-5 = 2x-6\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\) \( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x < 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{\left| {2x-1} \right| = 11-2x,}\\{\left| {2x-1} \right| = 2x-1\,\,\,\,\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {2x-1} \right| = 11-2x,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {2x-1} \right| = 2x-1.\,\,\,}\end{array}} \right.}\end{array}} \right.\) Рассмотрим первую систему последней совокупности: \(\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {2x-1} \right| = 11-2x}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{11-2x \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{2x-1 = 11-2x,}\\{2x-1 = 2x-11\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,}\\{x \le 5,5,}\\{x = 3,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 3.\) Рассмотрим вторую систему последней совокупности: \(\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {2x-1} \right| = 2x-1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{2x-1 = 2x-1,}\\{2x-1 = 1-2x}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \le 3,}\\{x \ge \dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left[ {\dfrac{1}{2};3} \right].\) Таким образом, решением исходного уравнения является \(x\, \in \,\left[ {\dfrac{1}{2};3} \right].\) Ответ: \(\left[ {\dfrac{1}{2};\;3} \right].\)