Решите неравенство...

$\dfrac{9}{x+1}- \dfrac{4}{x-4} \leq5\\ \dfrac{5(x+1)(x-4)-9(x-4)+4(x+1)}{(x-4)(x+1)}\geq0 \\ \dfrac{5(x^2-4x+x-4)-9x+36+4x+4}{(x+1)(x-4)}\geq0 \\ \dfrac{5x^2-20x+5x-20-9x+36+4x+4}{(x+1)(x-4)}\geq0\\ \dfrac{5x^2-20x+20}{(x-4)(x+1)}\geq0 \\ \boxed{\left \{ {{5x^2-20x+20\geq0} \atop {(x+1)(x-4)\ \textgreater \ 0}} \right. }\\\boxed{ \left \{ {{x^2-4x+4\geq0} \atop {(x-4)(x+1)\ \textgreater \ 0}} \right. }\\\boxed{ \left \{ {{(x-2)^2\geq0} \atop {(x+1)(x-4)\ \textgreater \ 0}} \right. }\\1)chisl\ nyl.:x=-1;x=4\\2)znam.\ nyl.:x=2$
ответ х∈(-∞;-1);{2};(4;∞)