Question

Помогите решить неравенство, плз.

(x - 3)(x^2 + 4)^(1/2) <= x^2 - 9

Answer 1

$(x-3)sqrt{x^2+4}leq (x-3)(x+3) \; (x-3)sqrt{x^2+4}-(x-3)(x+3) leq 0 \ (x-3)(sqrt{x^2+4}-x-3}leq 0 \ x = 3 ; sqrt{x^2+4}=x+3 \ x+3 textgreater 0; x textgreater -3 \ x^2+4-(x+3)^2=0 \ x=-frac{5}{6} \ -+- \ x in (-infty;-frac{5}{6}] cup [3;+infty)$