jmc

From the given inequality, either $\frac{2x-1}{x-1}>2$ or $\frac{2x-1}{x-1}<-2.$

The inequality $\frac{2x-1}{x-1}>2$ becomes $$\frac{2x-1}{x-1}-2>0,$$or $$\frac{1}{x-1}>0.$$This is satisfied when $x>1.$

The inequality $\frac{2x-1}{x-1}<-2$ becomes $$\frac{2x-1}{x-1}+2<0,$$or $$\frac{4x-3}{x-1}<0.$$If $x<\frac{3}{4},$ then $4x-3<0$ and $x-1<0,$ so the inequality is not satisfied.

If $\frac{3}{4}<x<1,$ then $4x-3>0$ and $x-1<0,$ so the inequality is satisfied.

If $x>1,$ then $4x-3>0$ and $x-1>0,$ so the inequality is not satisfied.

Thus, the solution is $$x\in \boxed{\left(\frac{3}{4},1\right)\cup (1,\infty )}.$$