Question

Решите неравенство |х²-8|≤2х

Answer 1

$|x^2-8|leq 2x\\Formyla:; ; ; |x| leq aquad Rightarrow quad -a leq x leq a ; ; .\\-2x leq x^2-8 leq 2xquad Rightarrow quad left { {{x^2-8 leq 2x} atop {x^2-8 geq -2x}} right. \\a); ; ; x^2-2x-8 leq 0; ,; ; x_1=-2,; ; x_2=4; ; (teorema; Vieta)\\(x+2)(x-4) leq 0qquad +++[-2]---[, 4,]+++\\xin [-2,4, ]\\b); ; x^2-8 geq -2x; ,; ; ; ; x^2+2x-8 geq 0; ,; ; x_1=-4,; x_2=2\\(x+4)(x-2) geq 0quad +++[-4]---[, 2, ]+++\\xin (-infty ,-4, ]cup [, 2,+infty )$

$c); ; left { {{xin [-2,4, ]} atop {xin (-infty ,-4, ]cup [, 2,+infty )}} right. quad Rightarrowquad underline {xin [, 2,4, ]}$