Помогите решить неравенство

$\frac{3^{|x|}*2^x-2^x-8*3^{|x|}+8}{2^{\sqrt{x}}-2} \geq 0\\\\ \frac{2^x*(3^{|x|}-1)-8*(3^{|x|}-1)}{2^{\sqrt{x}}-2} \geq 0\\\\ \frac{(2^x-8)*(3^{|x|}-1)}{2^{\sqrt{x}}-2} \geq 0\\\\$

если $x=0$ :
$\frac{(2^0-8)*(3^{0}-1)}{2^{0}-2} \geq 0\\\\ \frac{(1-8)*(1-1)}{1-2} \geq 0\\\\ 0 \geq 0$
Значит $x=0$ - часть решения
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если $x\neq0$ :
$\frac{(2^x-8)*(3^{|x|}-1)}{2^{\sqrt{x}}-2} \geq 0\\\\ \frac{2^x-8}{2^{\sqrt{x}}-2} \geq 0\\\\ \left \{ {{2^x-8 \geq 0} \atop {2^{\sqrt{x}}-2 \ \textgreater \ 0}} \right. \ or\ \left \{ {{2^x-8 \leq 0} \atop {2^{\sqrt{x}}-2 \ \textless \ 0}}$

$\left \{ {{x \geq 3} \atop {2^{\sqrt{x}} \ \textgreater \ 2}} \right. \ or\ \left \{ {{x \leq 3} \atop {2^{\sqrt{x}} \ \textless \ 2}}\\\\$

$\left \{ {{x \geq 3} \atop {\sqrt{x} \ \textgreater \ 1}} \right. \ or\ \left \{ {{x \leq 3} \atop {\sqrt{x} \ \textless \ 1}}$

$\left \{ {{x \geq 3} \atop {x \ \textgreater \ 1}} \right. \ or\ \left \{ {{x \leq 3} \atop { 0\leq x \ \textless \ 1}}$

$x \geq 3\ or\ 0\leq x \ \textless \ 1\\\\ x\in[0;\ 1)\cup[3;\ +\infty)$

второй и первый случаи вместе дают:
$x\in[0;\ 1)\cup[3;\ +\infty)$

Ответ: $[0;\ 1)\cup[3;\ +\infty)$