Помогииииитеее))))) 2,98 (б,в)

Решите задачу:

$\left \{ {{x^2 \leq 25} \atop { \frac{x^2+6x+9}{x^2-16} \leq 0 }} \right. \\\ \left \{ {{x\in[-5;5]} \atop { \frac{(x+3)^2}{(x-4)(x+4)} \leq 0 }} \right. \\\ \left \{ {{x\in[-5;5]} \atop { x\in(-4;4) }} \right. \\\ x\in(-4;4)$

$\left \{ {{(x-2)(x-3) \geq 0} \atop { \frac{x+3}{x^2-4} \geq 0}} \right. \\\ \left \{ {{x\in(2;3)} \atop { \frac{x+3}{(x-2)(x+2)} \geq 0}} \right. \\\ \left \{ {{x\in(-\infty;2]\cup[3;+\infty)} \atop {x\in[-3;-2)\cup(2\; +\infty)}} \right. \\\ x\in[-3;-2)\cup[3\; +\infty)$