\(\dfrac{2}{{x-2}}-\dfrac{4}{{x + 1}} \le 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{2}{{x-2}}-\dfrac{4}{{x + 1}}-1 \le 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{2x + 2-4x + 8-{x^2} + x + 2}}{{\left( {x-2} \right)\left( {x + 1} \right)}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2} + x-12}}{{\left( {x-2} \right)\left( {x + 1} \right)}} \ge 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\({x^2} + x-12 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -4,}\\{x = 3.\,\,\,\,}\end{array}} \right.\)
Найдём нули знаменателя:
\(\left( {x-2} \right)\left( {x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-2 = 0,}\\{x + 1 = 0\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,\,}\\{x = -1.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-4} \right] \cup \left( {-1;2} \right) \cup \left[ {3; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-4} \right] \cup \left( {-1;\;2} \right) \cup \left[ {3;\; + \infty } \right).\)