\(\dfrac{1}{{2-x}} + \dfrac{5}{{2 + x}} < 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{{2-x}} + \dfrac{5}{{2 + x}}-1 < 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{2 + x + 10-5x-4 + {x^2}}}{{\left( {2-x} \right)\left( {2 + x} \right)}} < 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2}-4x + 8}}{{\left( {x-2} \right)\left( {x + 2} \right)}} > 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\({x^2}-4x + 8 = 0;\,\,\,\,\,\,\,\,\,\,D = 16-32 = -16 < 0.\)
Найдём нули знаменателя:
\(\left( {x-2} \right)\left( {x + 2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-2 = 0,}\\{x + 2 = 0\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,}\\{x = -2.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-2} \right) \cup \left( {2; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-2} \right) \cup \left( {2;\; + \infty } \right).\)