Задача 9С. Решите уравнение \(\left| {\,{x^2}-2x-\left| {x + 1} \right|\,} \right| = \left| {{x^2} + x-3} \right|\)
ОТВЕТ: -1; 1/2; 2.
\(\left| {\,{x^2}-2x-\left| {x + 1} \right|\,} \right| = \left| {{x^2} + x-3} \right|.\) Уравнение вида \(\left| {f\left( x \right)} \right| = \left| {g\left( x \right)} \right|\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = g\left( x \right),\,\,\,}\\{f\left( x \right) = -g\left( x \right).}\end{array}} \right.\) \(\left[ {\begin{array}{*{20}{c}}{{x^2}-2x-\left| {x + 1} \right| = {x^2} + x-3,\,\,}\\{{x^2}-2x-\left| {x + 1} \right| = -{x^2}-x + 3}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {x + 1} \right| = 3-3x,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {x + 1} \right| = 2{x^2}-x-3.}\end{array}} \right.\) Рассмотрим первое уравнение: \(\left| {x + 1} \right| = 3-3x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x + 1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 1 = 3-3x,}\end{array}\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x + 1 < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{-x-1 = 3-3x}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \ge -1,}\\{x = 0,5,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < -1,}\\{x = 2\,\,\,\,}\end{array}\,} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \frac{1}{2}.\) Рассмотрим второе уравнение: \(\left| {x + 1} \right| = 2{x^2}-x-3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x + 1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x + 1 = 2{x^2}-x-3}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x + 1 < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{-x-1 = 2{x^2}-x-3}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \ge -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2{x^2}-2x-4 = 0,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < -1,\,\,\,\,\,\,\,\,\,\,}\\{2{x^2}-2 = 0}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \ge -1,}\\{\left[ {\begin{array}{*{20}{c}}{x = 2,}\\{x = -1}\end{array}} \right.}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x < -1}\\{\left[ {\begin{array}{*{20}{c}}{x = 1\,\,\,\,}\\{x = -1}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 2.\,\,}\end{array}} \right.\) Таким образом, решение исходного уравнения: \(x = -1,\,\,\,x = \frac{1}{2},\,\,\,\,x = 2.\) Ответ: \(-1;\;\;\frac{1}{2};\;\;2.\)