\(\dfrac{{\left| {x-3} \right|-\left| {x-2} \right|}}{{\left| {x + 1} \right| + x + 1}} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left| {x-3} \right|-\left| {x-2} \right| = 0,}\\{\left| {x + 1} \right| + x + 1 \ne 0.\,\,}\end{array}} \right.\)
\(\left| {x-3} \right|-\left| {x-2} \right| = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left| {x-3} \right| = \left| {x-2} \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-3 = x-2,\,\,}\\{x-3 = -x + 2}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 2,5.} \right.\)
Подставим \(x = 2,5\) в знаменатель исходного уравнения:
\(\left| {2,5 + 1} \right| + 2,5 + 1 = 7 \ne 0.\)
Следовательно, \(x = 2,5\) является решением исходного уравнения.
Ответ: 2,5.