|x^2+x-2|+|x+4|<=x^2+2x+6 — вопрос №343030

Ответ

Проверено экспертом

Автор - AnonimusPro

раскрываем модули и получаем систему:

$\left[begin{array}{cccc}left {begin{array}{ccc}x^2+x-2geq 0\x+4geq 0\x^2+x-2+x+4leq x^2+2x+6end{array}right.\left {begin{array}{ccc}x^2+x-2<0\x+4geq 0\-x^2-x+2+x+4leq x^2+2x+6end{array}right.\left {begin{array}{ccc}x^2+x-2< 0\x+4< 0\-x^2-x+2-x-4leq x^2+2x+6end{array}right.\ left {begin{array}{ccc}x^2+x-2geq 0\x+4<0\x^2+x-2-x-4leq x^2+2x+6end{array}right. end{array}right.$

$\left[begin{array}{cccc}left {begin{array}{ccc}x^2+x-2geq 0\x+4geq 0\2leq 6end{array}right.\left {begin{array}{ccc}x^2+x-2<0\x+4geq 0\x^2+xgeq 0end{array}right.\left {begin{array}{ccc}x^2+x-2< 0\x+4< 0\x^2+2x+4geq 0end{array}right.\ left {begin{array}{ccc}x^2+x-2geq 0\x+4<0\x+6geq 0end{array}right. end{array}right.$

решаем:

$1)left {begin{array}{ccc}x^2+x-2geq 0\x+4geq 0\2leq 6end{array}right.\x^2+x-2geq 0\D=1+8=9=3^2\x_1=frac{-1+3}{2}=1\x_2=frac{-1-3}{2}=-2$

+ - +

----[-2]------[1]------>x

$x in (-infty;-2]cup [1;+infty)$

$x+4geq 0\xgeq -4\xin [-4;+infty)\\2leq 6\x in R\left {begin{array}{ccc}x in (-infty;-2]cup [1;+infty)\xin [-4;+infty)\x in Rend{array}right. Rightarrow xin[-4;-2]cup [1;+infty)$

$2)left {begin{array}{ccc}x^2+x-2<0\x+4geq 0\x^2+xgeq 0end{array}right.\x^2+x-2<0\xin (-2;1)\xgeq -4\xin [-4;+infty)\x(x+1)geq 0\x_1=0\x_2=-1$

+ - +

----[-1]-----[0]----->x

$xin (-infty;-1]cup [0;+infty)$

$left {begin{array}{ccc}xin (-2;1)\xin [-4;+infty)\xin (-infty;-1]cup [0;+infty)end{array}right. Rightarrow x in(-2;-1]cup [0;1)$

$3)left {begin{array}{ccc}x^2+x-2< 0\x+4< 0\x^2+2x+4geq 0end{array}right.\x^2+x-2<0\xin (-2;1)\x+4<0\xin (-infty;-4)\x^2+2x+4geq 0\left { {{D=4-16<0} atop {a>0}} right. Rightarrow x^2+2x+4>0, forall x in R\x in R\left {begin{array}{ccc}xin (-2;1)\xin (-infty;-4)\xin Rend{array}right. Rightarrow x in varnothing$

$4)left {begin{array}{ccc}x^2+x-2geq 0\x+4<0\x+6geq 0end{array}right.\x^2+x-2geq 0\x in (-infty;-2]cup [1;+infty)\x+4<0\xin (-infty;-4)\x+6geq 0\x in[-6;+infty)\left {begin{array}{ccc}x in (-infty;-2]cup [1;+infty)\xin (-infty;-4)\x in[-6;+infty)end{array}right. Rightarrow x in[-6;-4)$

В итоге:

$left[begin{array}{cccc}xin[-4;-2]cup [1;+infty)\x in(-2;-1]cup [0;1)\x in varnothing\ x in[-6;-4) end{array}right. Rightarrow x in[-6;-1]cup [0;+infty)$

Ответ: $x in[-6;-1]cup [0;+infty)$