Question

Помогите пожалуйста.

Answer 1

$\left \{ {{(x+3)(x-4) \leq x^2} \atop { \frac{x+1}{3}- \frac{x}{4} \ \textgreater \ 0}} \right. \\ \\ \left \{ {{x^2+3x-4x-12 \leq x^2} \atop { \frac{4(x+1)-3x}{12}\ \textgreater \ 0}} \right. \\ \\ \left \{ {{x^2-x^2-4x-12 \leq 0} \atop {4x++4-3x\ \textgreater \ 0}} \right. \\ \\ \left \{ {{-4x \leq 12} \atop {x\ \textgreater \ -4}} \right. \\ \left \{ {{x \leq -3} \atop {x\ \textgreater \ -4}} \right.$

ответом будет пересечение двух множеств ]-∞;-3] и ]-4;+∞[