Решите систему неравенств под буквой Б

[latex] \left \{ {{x^2-x-12\ \textless \ 0,} \atop {x^2-4x+3 \geq 0;}} \right. \left \{ {{(x+3)(x-4)\ \textless \ 0,} \atop {(x-1)(x-3)\geq 0;}} \right. \left \{ {{-3\ \textless \ x\ \textless \ 4,} \atop {\left \{ {{x \leq 1,} \atop {x \geq 3;}} \right.}} \right. \left [ {{-3\ \textless \ x \leq 1,} \atop {3 \leq x\ \textless \ 4;}} \right. \\ x\in(-3;1]\cup[3;4).[/latex] [latex] \left \{ {{x^2+x-2 \geq 0,} \atop {\frac{x+2}{x-4}\leq 0}} \right. \left \{ \begin{array}{ccc}(x+2)(x-1) \geq 0,\\x-4 \neq 0,\\(x+2)(x-4) \leq 0;\end{array}\right. \left \{ \begin{array}{ccc} \left [ {{x \leq -2,} \atop {x \geq 1,}} \right. \\x \neq 4,\\-2 \leq x \leq 4;\end{array}\right. \left [ {{x=-2,} \atop {1 \leq x\ \textless \ 4;}} \right. \\ x\in\{-2\}\cup[1;4).[/latex] [latex] \left \{ {{|x| \leq 2,} \atop {x^2-x-6 \geq 0;}} \right. \left \{ {{-2 \leq x \leq 2,} \atop {(x+2)(x-3) \geq 0;}} \right. \left \{ {{-2 \leq x \leq 2,} \atop {(x+2)(x-3) \geq 0;}} \right. \left \{ {{-2 \leq x \leq 2,} \atop { \left [ {{x \leq -2,} \atop {x \geq 3;}} \right. }} \right. \\ x=-2.[/latex]