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