$\left| 2x - 1\right| \mid 2x - 1 > 0\, \iff x > \frac{1}{2}$:
\[2x - 1 > \frac{1}{\sqrt{x}} \implies 4x^3 -4x^2 + x - 1 > 0 \implies 4x^2\left(x - 1\right) + \left(x - 1\right) > 0 \implies \left(x - 1\right) \cdot \left(4x^2 + 1\right) > 0 \iff x > 1\,.\]
$\left| 2x - 1\right| \mid 2x - 1 < 0\, \iff x < \frac{1}{2}$:
\[-2x + 1 > \frac{1}{\sqrt{x}} \implies 4x^3 -4x^2 + x - 1 > 0 \implies 4x^2\left(x - 1\right) + \left(x - 1\right) > 0 \implies \left(x - 1\right) \cdot \left(4x^2 + 1\right) > 0 \iff x > 1\,.\]
Dunque non esiste soluzione nel caso $\left| 2x - 1\right| \mid 2x - 1 < 0\,$ in quanto $x < \frac{1}{2} \land x > 1 = \emptyset\,$ e $\,2x - 1 > 0 \iff x > 1\,$.