\(\dfrac{{x + 4}}{{x-2}} \le \dfrac{2}{{x + 1}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{x + 4}}{{x-2}}-\dfrac{2}{{x + 1}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{{x^2} + 5x + 4-2x + 4}}{{\left( {x-2} \right)\left( {x + 1} \right)}} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{{x^2} + 3x + 8}}{{\left( {x-2} \right)\left( {x + 1} \right)}} \le 0.\)
Решим неравенство методом интервалов. Найдём нули числителя:
\({x^2} + 3x + 8 = 0;\,\,\,\,\,\,\,\,\,\,D = 9-32 = -23 < 0.\)
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: \(x\, \in \,\left( {-1;2} \right).\)
Ответ: \(\left( {-1;\;2} \right).\)