Помогите, пожалуйста, с 7 и 8!!

Question 1

Question 2

Решите задачу:

$7)\; \; \frac{5-|x|}{x^2+|x|-2} \geq \frac{|x|-5}{x^2-1}\\\\t=|x| \geq 0\; ,\; \; x^2=|x|^2=t^2\\\\ \frac{5-t}{t^2+t-2} \geq \frac{t-5}{t^2-1}\\\\ \frac{5-t}{(t-1)(t+2)}- \frac{t-5}{(t-1)(t+1)} \geq 0\\\\ \frac{(5-t)(t+1)-(t-5)(t-1)}{(t-1)(t+2)(t+1)} \geq 0\\\\ \frac{4t+5-t^2-(t^2-6t+5)}{(t-1)(t+2)(t+1)} \geq 0\; ,\; \; \frac{-2t^2 +10t}{(t-1)(t+2)(t+1)} \geq 0\; ,\; \; \frac{-2t(t-5)}{(t-1)(t+2)(t+1)} \geq 0\\\\ \frac{t(t-5)}{(t-1)(t+2)(t+1)} \leq 0\\\\---(-2)+++(-1)--[\, 0\, ]++(1)---[\, 5\, ]+++$

$t\in (-\infty ,-2)\cup (-1,0\, ]\cup (1,5\, ]\\\\t \geq 0\; \; \Rightarrow \; \; \; |x|\in \{0\}\cup (1,5\, ]\; \; \Rightarrow \; \; \left \{ {{|x|=0,\; |x| \leq 5} \atop {|x|\ \textgreater \ 1}} \right. \; ,\; \left \{ {{x=0,\; -5 \leq x \leq 5} \atop {x\ \textgreater \ 1\; ili\; x\ \textless \ -1}} \right. \\\\x\in [-5,-1)\cup (1,5\, ]$

$8)\; \; \frac{x^2+|x|-2}{x^2+|x|-6}\ \textgreater \ 0\\\\t=|x| \geq 0\; ,\; \; x^2=|x|^2=t^2\\\\ \frac{t^2+t-2}{t^2+t-6} \ \textgreater \ 0\; ,\; \; \frac{(t-1)(t+2)}{(t-2)(t+3)}\ \textgreater \ 0\\\\+++(-3)---(-2)+++(1)---(2)+++\\\\t\in (-\infty -3)\cup (-1,1)\cup (2,+\infty )\\\\t \geq 0\; \; \Rightarrow \; \; t\in (0,1)\cup (2,+\infty )\; \; \Rightarrow \\\\ \left \{ {{0\ \textless \ |x|\ \textless \ 1} \atop {|x|\ \textgreater \ 2}} \right. \; ,\; \; \left \{ {{-1\ \textless \ x\ \textless \ 1} \atop {x\ \textgreater \ 2\; ili\; x\ \textless \ -2}} \right. \; \; \Rightarrow \; \; \; x\in \varnothing$