Уравнения с модулями. Задача 20

ОТВЕТ: \( \pm 1;\;\; \pm 2.\)

\(\left| {\,2{x^2} + x-4\,} \right| = \left| {\,{x^2} + 2x-2\,} \right|.\)

Уравнение вида  \(\left| {f\left( x \right)} \right| = \left| {g\left( x \right)} \right|\)  равносильно совокупности:  \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = g\left( x \right),\;\,}\\{f\left( x \right) = -g\left( x \right).}\end{array}} \right.\)

\(\left| {\,2{x^2} + x-4\,} \right| = \left| {\,{x^2} + 2x-2\,} \right|\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{2{x^2} + x-4 = {x^2} + 2x-2,\;}\\{2{x^2} + x-4 = -{x^2}-2x + 2}\end{array}} \right.\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{{x^2}-x-2 = 0,}\\{{x^2} + x-2 = 0\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\left( {x-2} \right)\left( {x + 1} \right) = 0,}\\{\left( {x-1} \right)\left( {x + 2} \right) = 0\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 2,\;\,}\\{x = -1,}\\{x = 1,\;\;\,}\\{x = -2.}\end{array}} \right.\)

Ответ:   \( \pm 1;\;\;\; \pm 2.\)