Question

Помогите найти целые решения неравенства. |x-7|(36-x^2)≥0

Answer 1

$|x-7|(36-x^2) \geq 0\\\\|x-7| \geq 0\; \; pri\; \; x\in (-\infty,+\infty)\; \; \to \; \; 36-x^2 \geq 0\\\\x^2-36 \leq 0\; \; \to \; \; (x-6)(x+6) \leq 0\; \; \to \; \; x\in [\, -6,6\, ]\\\\|x-7|=0\; \; pri\; \; x=7\\\\x\in [\, -6,6\, ]U\{7\}\\\\Celue\; resheniya:\; \; -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7.$