Solve the following

Solution:

As, $\frac{1}{|x|-3}<\frac{1}{2}$

$\Rightarrow \frac{1}{|x|-3}-\frac{1}{2}<0$

$\Rightarrow \frac{2-(|x|-3)}{2(|x|-3)}<0$

$\Rightarrow \frac{2-|x|+3}{|x|-3}<0$

$\Rightarrow \frac{5-|x|}{|x|-3}<0$

Case I: When $x \geq 0,|x|=x$

$\frac{5-x}{x-3}<0$

$\Rightarrow(5-x<0$ and $x-3>0)$ or $(5-x>0$ and $x-3<0)$

$\Rightarrow(x>5$ and $x>3)$ or $(x<5$ and $x<3)$

$\Rightarrow x>5$ and $x<3$

$\Rightarrow x \in[0,3) \cup(5, \infty)$

Case II: When $x \leq 0,|x|=-x$,

$\frac{5+x}{-x-3}<0$

$\Rightarrow \frac{x+5}{x+3}>0$

$\Rightarrow(x+5>0$ and $x+3>0)$ or $(x+5<0$ and $x+3<0)$

$\Rightarrow(x>-5$ and $x>-3)$ or $(x<-5$ and $x<-3)$

$\Rightarrow x>-3$ and $x<-5$

$\Rightarrow x \in(-\infty,-5) \cup(-3,0]$

So, from both the cases, we get

$x \in(-\infty,-5) \cup(-3,0] \cup[0,3) \cup(5, \infty)$

$\therefore x \in(-\infty,-5) \cup(-3,3) \cup(5, \infty)$