\(\left| {\dfrac{{x-1}}{{2x + 1}}} \right| = \left| {\dfrac{{2x + 1}}{{x-1}}} \right|.\)
Уравнение вида \(\left| {f\left( x \right)} \right| = \left| {g\left( x \right)} \right|\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = g\left( x \right),\;\,}\\{f\left( x \right) = -g\left( x \right).}\end{array}} \right.\)
\(\left| {\dfrac{{x-1}}{{2x + 1}}} \right| = \left| {\dfrac{{2x + 1}}{{x-1}}} \right|\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{{x-1}}{{2x + 1}} = \dfrac{{2x + 1}}{{x-1}},\;}\\{\dfrac{{x-1}}{{2x + 1}} = -\dfrac{{2x + 1}}{{x-1}}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{2x + 1 \ne 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,}\\{x-1 \ne 0,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{array}}\\{\left[ {\begin{array}{*{20}{c}}{{{\left( {x-1} \right)}^2}-{{\left( {2x + 1} \right)}^2} = 0,}\\{{{\left( {x-1} \right)}^2} + {{\left( {2x + 1} \right)}^2} = 0\,}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{x \ne -0,5,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,}\\{x \ne 1,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\;\;\;\;\;}\end{array}}\\{\left[ {\begin{array}{*{20}{c}}{\left( {x-1-2x-1} \right)\left( {x-1 + 2x + 1} \right) = 0,}\\{{x^2}-2x + 1 + 4{x^2} + 4x + 1 = 0\;\;\;\;\;\;\,\,}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{x \ne -0,5,}\\{x \ne 1,\,\;\;\;\;\;}\end{array}}\\{\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 0,\;\;}\\{x \notin R\;\;}\end{array}\;\,} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -2,}\\{x = 0.\;\;}\end{array}} \right.\)
Ответ: \(-2;\;\;\;0.\)