$$\frac{|2x-4|}{x+3} \le 0$$
We can easily see that if $x=2$, the inequality will become zero, as the numerator also cancels out to zero. This is a solution, but this isn't the end - we're also looking for negative solutions.
So if we know that $x=2$ produces $0$, then anything less than $x=2$ will produce a numerator that is negative - however, this isn't possible since there are absolute value signs around the numerator. This means that the numerator will never be negative - so we look at the denominator ($x+3$).
We're looking for values of $x+3$ that are less than or equal to zero. Let's set up an equation:
$x+3 \le 0$
We can easily solve this to $x \le -3$, but this is wrong. We have to remember the fact that since this is the denominator of a fraction, it cannot equal zero lest the value be undefined. We can see this happen when $x = - 3$, so $ x \neq 3 $ so we remove that from $x \le - 3$, leaving us with $x < - 3$.
Final answer: $x < -3, x =2$.
Commonly when solving equations with absolute values in them a method is to square both sides, making the absolute value signs disappear. In this case:
$$\frac{(2x-4)^2}{(x+3)^2} \le 0$$
One should try and avoid this outcome unless one knows for sure that both sides are not negative because in doing so, half of the solution is eliminated.