\(\dfrac{{{x^2}-x}}{{{x^2}-x + 1}}-\dfrac{{{x^2}-x + 2}}{{{x^2}-x-2}} = 1.\)
Пусть \({x^2}-x = t.\) Тогда исходное уравнение примет вид:
\(\dfrac{t}{{t + 1}}-\dfrac{{t + 2}}{{t-2}} = 1\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{{t^2}-2t-{t^2}-2t-t-2-{t^2} + 2t-t + 2}}{{\left( {t-2} \right)\left( {t + 1} \right)}} = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\dfrac{{-{t^2}-4t}}{{\left( {t-2} \right)\left( {t + 1} \right)}} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\dfrac{{-t\left( {t + 4} \right)}}{{\left( {t-2} \right)\left( {t + 1} \right)}} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{t \ne 2,\;\;\,}\\{t \ne -1,\,}\end{array}}\\{\left[ {\begin{array}{*{20}{c}}{t = 0,\;}\\{t = -4}\end{array}} \right.}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{t = 0,\;\;}\\{t = -4.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{{x^2}-x = 0,\;}\\{{x^2}-x = -4}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x\left( {x-1} \right) = 0,\,\;}\\{{x^2}-x + 4 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 1,}\\{x = 0,}\\{x \notin R}\end{array}} \right.\;\;\;\; \Leftrightarrow \,\;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = 1,\,}\\{x = 0.}\end{array}} \right.\)
Ответ: \(0;\;\;\;\;1.\)