Answer
$(0,0.5]\cup(1,3]$
Work Step by Step
1. Rewrite inequality as $2-\frac{1}{x-1}-\frac{3}{x}\leq0, \frac{2x^2-2x-2x-3x+3}{x(x-1)}= \frac{2x^2-7x+3}{x(x-1)}$
$f(x)=\frac{(2x-1)(x-3)}{x(x-1)}\leq0$
2 Cut points $0, 1/2, 1, 3$, build up a sign table as shown.
3. The solution is $(0,0.5]\cup(1,3]$
