\(\dfrac{1}{{x-1}} \ge -\dfrac{2}{{x + 2}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{{x-1}} + \dfrac{2}{{x + 2}} \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{3x}}{{\left( {x-1} \right)\left( {x + 2} \right)}} \ge 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\(3x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = 0.\)
Найдём нули знаменателя:
\(\left( {x-1} \right)\left( {x + 2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-1 = 0,}\\{x + 2 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,}\\{x = -2.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-2;0} \right] \cup \left( {1; + \infty } \right).\)
Ответ: \(\left( {-2;\;0} \right] \cup \left( {1;\; + \infty } \right).\)