<=       (1)

 -  <= 

 -  <=    (written with the common denominator 2x )

 <= 

 <=      (2)


On the left side, we have a rational function of the three factors (2-x), (2+x) and (2x).


There are three critical values, where the factor become equal to zero and change their sign: -2, 0, and 2.


And there are four interval to analyse:  a) x < -2;  b) -2 <= x < 0;  c) 0 < x < 2;  and  d) x >= 2.


In the interval  a) x < - 2  factors (x+2) and (2x) are negative;  factor (2-x) is positive.  So inequality (2) is FALSE.


In the interval  b) -2 <= x < 0  the factor (x+2) is positive, factor (2x) is negative; factor (2-x) is positive.  
So, the whole function in the left side of (2) is negative, and the inequality (2) IS TRUE.


In the interval  c) 0 < x < 2  the factors (x+2)  and (2x)  are positive;  the factor (x-2) is positive; so inequality (2) is FALSE.


In the interval  d) x >= 2  factors (x+2) and (2x) are positive;  factor (2-x) is negative.  So, the whole function in the left side of (2) is negative, and the inequality (2) IS TRUE.


Thus the inequality (2) is TRUE in these two intervals

     -2 <= x < 0    and    x >= 2.


Since inequalities (1) and (2) are EQUIVALENT (!),  your answer is:


Answer.  The original inequality is true  at  -2 <= x < 0    and    x >= 2.

         The solution is the set  [,)  U  [,].