Question

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

Answer 1

$3^{|x|}-8- frac{3^{|x|}+9}{9^{|x|}-4cdot 3^{|x|}+3} leq frac{5}{3^{|x|}-1} \\ODZ: left { {{3^{|x|}-1ne 0} atop {9^{|x|}-4cdot 3^{|x|}+3ne 0}} right. ; left { {{3^{|x|}ne 1} atop {(3^{|x|}-1)(3^{|x|}-3)ne 0}} right. ; left { {{xne 0} atop {|x|ne 0; ,; |x|ne1}} right. ; left { {{xne 0}atop {xne pm 1}} right. \\t=3^{|x|} textgreater 0; ,; 9^{|x|}=(3^{|x|})^2; ,; ; ; ; t-8-frac{t-9}{(t-1)(t-3)}-frac{5}{t-1} leq 0$

$frac{(t-8)(t-1)(t-3)-(t+9)-5(t-3)}{(t-1)(t-3)}leq 0$

$frac{(t-3)(t^2-9t+8-5)-t-9}{(t-1)(t-3)} leq 0\\ frac{t^3-12t^2+29t-18}{(t-1)(t-3)} leq 0; ,\\ t^3-12t^2+29t-18=(t-1)(t^2-11t+18)=(t-1)(t-2)(t-9)\\frac{(t-1)(t-2)(t-9)}{(t-1)(t-3)} leq 0; ,; ; ; frac{(t-2)(t-9)}{t-3} leq 0; ,; ; (t textgreater 0; ,; tne pm 1)\\(0)---(1)---[2]+++(3)---[9]+++\\tin (0,1)cup (1,2; ]cup (3,9, ]$

$1); ; 0 textless 3^{|x|} textless 1; ; to ; ; ; 3^{|x|} textless 3^0; ; to ; ; |x| textless 0; net; resheniij\\2); ; 1 textless 3^{|x|} leq 2; ; to ; ; 3^0 textless 3^{|x|} leq 3^{log_32}; ; to \\|x| leq log_32; ; (log_32 textless log_33=1; ,; ; log_32approx 0,63)\\-log_32 leq x leq log_32\\xin [-log_32, ;, log_32, ]$

$3); ; 3 textless 3^{|x|} leq 9; ; to ; ; 3^1 textless 3^{|x|} leq 3^2; ; to ; ; 1 textless |x| leq 2\\ left { {{|x| textgreater 1} atop {|x| leq 2}} right. ; ,; ; left { {{ left [ {{x textgreater 1} atop {x textless -1}} right. } atop { -2 leq x leq 2} right.$

$left { {{xin (-infty ,-1)cup (1,+infty )} atop {xin [-2,2, ]}} right. ; ; Rightarrow ; ; ; xin [-2,-1)cup (1,2, ]\\Otvet:; ; xin [-2,-1)cup [-log_32, ;, log_32]cup (1,2, ]; .$