I'm a bit rusty, but here you go...
|5/2x-1| < 1
2x - 1 <0 and 2x - 1 >5 [because dividing by 0 gives infinity, dividing by 5 gives 1, and anything in between will push it above 1]
∴ x< 1/2 and x >3
(2-x)(2x-1)(x+3) less than/equal to 0
to be equal to 0, x must be either 2, 1/2 or -3
a quick...