\(\left| {{x^2}-x-1} \right| = {x^2} + 2x + 1.\)
Уравнение вида \(\left| {f\left( x \right)} \right| = g\left( x \right)\) равносильно системе: \(\left\{ {\begin{array}{*{20}{c}}{g\left( x \right) \ge 0,\;\;\;\;\;\;\;\,\;\;}\\{\left[ {\begin{array}{*{20}{c}}{f\left( x \right) = g\left( x \right),\;\,}\\{f\left( x \right) = -g\left( x \right).}\end{array}} \right.}\end{array}} \right.\)
\(\left| {{x^2}-x-1} \right| = {x^2} + 2x + 1\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{{x^2} + 2x + 1 \ge 0,\;\;\;\;\;\;\;\,\;\,\;\,\;\,\;}\\{\left[ {\begin{array}{*{20}{c}}{{x^2}-x-1 = {x^2} + 2x + 1,\,}\\{{x^2}-x-1 = -{x^2}-2x-1}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{{{\left( {x + 1} \right)}^2} \ge 0,\,\;\;\;}\\{\left[ {\begin{array}{*{20}{c}}{3x = -2,\;\;\;\;\;\;\,\,}\\{x\left( {2x + 1} \right) = 0}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{x \in R,\;\;\;\;}\\{\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{2}{3},}\\{\begin{array}{*{20}{c}}{x = -\dfrac{1}{2},}\\{x = 0\;\;\;\;}\end{array}}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = -\dfrac{2}{3},}\\{\begin{array}{*{20}{c}}{x = -\dfrac{1}{2},}\\{x = 0.\,\;\;\;}\end{array}}\end{array}} \right.\)
Ответ: \(-\dfrac{2}{3};\;\;\;-\dfrac{1}{2};\;\;\;0.\)