Question

Решите неравенство с модулями

Answer 1

Ответ:

решение представлено на фото

,

Answer 2

Пусть $3^{|x|}=t\geq 1$, тогда $9^{|x|}=t^2$

$t-8-\frac{t+9}{t^2-4t+3}\leq \frac{5}{t-1}\\t-8-\frac{t+9}{(t-1)(t-3)}-\frac{5(t-3)}{(t-1)(t-3)}\leq 0\\t-8-\frac{t+9+5t-15}{(t-1)(t-3)}\leq 0\\t-8-\frac{6t-6}{(t-1)(t-3)}\leq 0\\\left \{ {{\frac{(t-8)(t-3)}{t-3}-\frac{6}{t-3}\leq 0} \atop {t\neq 1}} \right. \\\left \{ {{\frac{t^2-11t+18}{t-3}\leq 0} \atop {t\neq 1}} \right. \\\left \{ {{\frac{(t-2)(t-9)}{t-3}\leq 0} \atop {t\neq 1}} \right. \Rightarrow t\in(-\infty; 1)\cup(1;2]\cup(3;9],\ t\geq 1\Rightarrow t\in(1;2]\cup(3;9]$

$\left [ {{1<3^{|x|}\leq 2} \atop {3<3^{|x|}\leq 9}} \right. \\\left [ {{0<|x|\leq \log_3{2}} \atop {1<|x|\leq 2}} \right. \Rightarrow x\in [-2;-1)\cup[-\log_3{2};0)\cup(0;\log_3{2}]\cup(1;2]$

Ответ: $[-2;-1)\cup[-\log_3{2};0)\cup(0;\log_3{2}]\cup(1;2]$