15 помогите, пожалуйста Неравенство с модулями Не понимаю, как это делать

[latex]3^{|x|}-8- \frac{3^{|x|}+9}{9^{|x|}-4\cdot 3^{|x|}+3} \leq \frac{5}{3^{|x|}-1} \\\\ODZ: \left \{ {{3^{|x|}-1\ne 0} \atop {9^{|x|}-4\cdot 3^{|x|}+3\ne 0}} \right. \; \left \{ {{3^{|x|}\ne 1} \atop {(3^{|x|}-1)(3^{|x|}-3)\ne 0}} \right. \; \left \{ {{x\ne 0} \atop {|x|\ne 0\; ,\; |x|\ne1}} \right. \; \left \{ {{x\ne 0}\atop {x\ne \pm 1}} \right. \\\\t=3^{|x|}\ \textgreater \ 0\; ,\; 9^{|x|}=(3^{|x|})^2\; ,\; \; \; \; t-8-\frac{t-9}{(t-1)(t-3)}-\frac{5}{t-1} \leq 0[/latex] [latex]\frac{(t-8)(t-1)(t-3)-(t+9)-5(t-3)}{(t-1)(t-3)}\leq 0[/latex] [latex] \frac{(t-3)(t^2-9t+8-5)-t-9}{(t-1)(t-3)} \leq 0\\\\ \frac{t^3-12t^2+29t-18}{(t-1)(t-3)} \leq 0\; ,\\\\ t^3-12t^2+29t-18=(t-1)(t^2-11t+18)=(t-1)(t-2)(t-9)\\\\\frac{(t-1)(t-2)(t-9)}{(t-1)(t-3)} \leq 0\; ,\; \; \; \frac{(t-2)(t-9)}{t-3} \leq 0\; ,\; \; (t\ \textgreater \ 0\; ,\; t\ne \pm 1)\\\\(0)---(1)---[2]+++(3)---[9]+++\\\\t\in (0,1)\cup (1,2\; ]\cup (3,9\, ][/latex] [latex]1)\; \; 0\ \textless \ 3^{|x|}\ \textless \ 1\; \; \to \; \; \; 3^{|x|}\ \textless \ 3^0\; \; \to \; \; |x|\ \textless \ 0\; net\; resheniij\\\\2)\; \; 1\ \textless \ 3^{|x|} \leq 2\; \; \to \; \; 3^0\ \textless \ 3^{|x|} \leq 3^{log_32}\; \; \to \\\\|x| \leq log_32\; \; (log_32\ \textless \ log_33=1\; ,\; \; log_32\approx 0,63)\\\\-log_32 \leq x \leq log_32\\\\x\in [-log_32\, ;\, log_32\, ][/latex] [latex]3)\; \; 3\ \textless \ 3^{|x|} \leq 9\; \; \to \; \; 3^1\ \textless \ 3^{|x|} \leq 3^2\; \; \to \; \; 1\ \textless \ |x| \leq 2\\\\ \left \{ {{|x|\ \textgreater \ 1} \atop {|x| \leq 2}} \right. \; ,\; \; \left \{ {{ \left [ {{x\ \textgreater \ 1} \atop {x\ \textless \ -1}} \right. } \atop { -2 \leq x \leq 2} \right. [/latex] [latex] \left \{ {{x\in (-\infty ,-1)\cup (1,+\infty )} \atop {x\in [-2,2\, ]}} \right. \; \; \Rightarrow \; \; \; x\in [-2,-1)\cup (1,2\, ]\\\\Otvet:\; \; x\in [-2,-1)\cup [-\log_32\, ;\, \log_32]\cup (1,2\, ]\; .[/latex]