\(\dfrac{{\left| {x-1} \right|}}{{x-1}} + \dfrac{{\left| {x + 4} \right|}}{{x + 4}} = -2.\)
Решим исходное уравнение методом интервалов:
\(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < -4,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {x-1} \right)}}{{x-1}}-\dfrac{{x + 4}}{{x + 4}} = -2,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{-4 < x < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {x-1} \right)}}{{x-1}} + \dfrac{{x + 4}}{{x + 4}} = -2,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{x-1}}{{x-1}} + \dfrac{{x + 4}}{{x + 4}} = -2\,}\end{array}\,\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < -4,\,\,}\\{-2 = -2}\end{array}\,\,\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{-4 < x < 1,}\\{0 = -2\,\,\,\,\,\,\,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x > 1,\,\,\,\,}\\{2 = -2.}\end{array}\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;-4} \right).\)
Ответ: \(\left( {-\infty ;\;-4} \right).\)