Find the range of values of $\displaystyle x$ for which $\displaystyle |\frac{2x+5}{x^2-4}|\geqslant \frac{1}{5}$
So you want to find x so that $\displaystyle \frac{2x+ 5}{x^2- 4}\ge \frac{1}{5}$ or $\displaystyle \frac{2x+ 5}{x^2- 4}\le -\frac{1}{5}$
If |x|> 2, then $\displaystyle x^2- 4> 0$ so those are the same as
$\displaystyle 5(2x+ 5)= 10x+ 25> x^2- 4$ and
$\displaystyle 5(2x+ 5)= 10x+ 25> 4- x^2$
which are, of course, the same as
$\displaystyle x^2- 10x- 29> 0$ and
$\displaystyle x^2+ 10x+ 21> 0$
What values of x> 2 or x< -2 satisfy those?
If |x|< 2 then $\displaystyle x^2- 4< 0$ so those are the same as
$\displaystyle 5(2x+ 5)= 10x+ 25< x^2- 4$ and
$\displaystyle 5(2x+ 5)= 10x+ 25< 4- x^2$ which are the same as
$\displaystyle x^2- 10x- 29< 0$ and
$\displaystyle x^2+ 10x+ 21< 0$
What values of x, -2< x< 2, satisfy those?