\(\dfrac{{\left| {x + 1} \right|}}{{x + 1}} + \dfrac{{\left| {x + 3} \right|}}{{x + 3}} = 0.\)
Решим исходное уравнение методом интервалов:
\(\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < -3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {x + 1} \right)}}{{x + 1}}-\dfrac{{x + 3}}{{x + 3}} = 0,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{-3 < x < -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{-\left( {x + 1} \right)}}{{x + 1}} + \dfrac{{x + 3}}{{x + 3}} = 0,\,\,\,\,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x > -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{x + 1}}{{x + 1}} + \dfrac{{x + 3}}{{x + 3}} = 0\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x < -3,\,\,}\\{-2 = 0\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{-3 < x < -1,}\\{0 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x > -1,\,\,\,\,}\\{2 = 0.\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-3;-1} \right).\,\)
Ответ: \(\left( {-3;\;-1} \right).\)