Question

Решить неравенство :
| х² - 2х - 15 | ≤ - 2х + 15

Answer 1

$|x^2-2x-15| leq -2x+15 \ x^2-2x-15V0 \ D = 4+60= 64 \ x_{1,2} = frac{2б8}{2} \ x_1 = 5; x_2 = -3 \ |(x-5)(x+3)| leq -2x+15$
Переносим в совокупность:
$(x-5)(x+3) leq -2x+15 \ x>5; x<-3 \ x^2-2x-15+2x-15 leq 0 \ x^2 leq 30 \ x leq бsqrt{30} \ x in [-sqrt{30};sqrt{30}] \ -3<x<5 \ -x^2+2x+15 leq -2x+15 \ -x^2+4x leq 0 \ x(4-x) leq 0 \ x =0; x = 4;  \ x in (-infty;0] cup [4;+infty) \ x in [4;sqrt{30}] cup [-sqrt{30};0]$