I need help understanding a simple concept that I can't seem to get with a certain type of inequality I've got this inequality: 
$$\dfrac{1}{x}-\dfrac{x}{2x-1}\geq1$$
And the solutions are supposed to be $(0,1/2)$. But why? Whenever I make my inequality table, it ends up telling me the solution is $(-∞,0) \cup (1/2,+∞)$
Can someone help me understand what is going on? I really need to know this to study for a test. 
EDIT: Alright, to make it clearer this is what I've got after simplifying everything: 
$$\dfrac{\left(-3x^2+3x-1\right)}{x\left(2x-1\right)}>=0$$
After that I just put all the values on a table, in which I look for the numbers that make 2x-1 and x equal 0. The polynomial in the numerator can never be zero since its discriminant is less than 0. That's basically where I'm stuck at. I would post the table, but I've no clue where I could create one, I'm doing this on paper. 
 A: $\dfrac{1}{x} - \dfrac{x}{2x-1} \geq 1 \iff \dfrac{1}{x} - \dfrac{x}{2x-1} - 1 \geq 0 \iff \dfrac{2x-1 - x^2 - x(2x - 1)}{x(2x - 1)} \geq 0 \iff \dfrac{-3x^2 + 3x - 1}{x(2x - 1)} \geq 0$. 
Observe that: $-3x^2 + 3x  - 1 < 0$, $\forall x$ because $\triangle = 3^2 - 4(-3)(-1) = 9 - 12 = -3 < 0$, and $-3 < 0$. Thus the inequality is equivalent to:
$x(2x - 1) < 0$, and this gives: $0 < x < \dfrac{1}{2}$
A: The inequality is equivalent to 
$$
\begin{array}{rclr}
  \frac{2x-1 - x^2}{x(2x-1)} -1       &\geq& 0  & \iff \\
  \frac{2x-1 - x^2-2x^2 + x}{x(2x-1)} &\geq& 0  & \iff \\
  \frac{-3x^2 + 3x -1 }{x(2x-1)}      &\geq& 0. & \ \\
\end{array}
$$
The numerator is a negative term (the polynomial opens downwards and has no real roots) thus the quotient is non-negative precisely when $x(2x-1) < 0$, i.e. 
when $x \in (0,1/2)$.
A: Combine the fractions on the LHS of the inequality by finding the common denominator and note that the numerator is then the square of $x-1$:
$$\frac{1}{x}-\frac{x}{2x-1}=\frac{2x-1-x^2}{x(2x-1)}=-\frac{(x-1)^2}{x(2x-1)}=\frac{(x-1)^2}{x(1-2x)}.$$
Assuming $x\neq 1$, we divide both sides by $(x-1)^2$:
$$1\leq \frac{(x-1)^2}{x(1-2x)} \implies \frac{1}{(x-1)^2}\leq \frac{1}{x(1-2x)}$$
This tells us that $1/x$ and $1/(1-2x)$ must either be both positive or both negative, and hence that $x$ and $1-2x$ must either be both positive or negative as well. All that's left now is some very simple case work.
