algebra precalculus - Absolute value fraction problem

Is there an easier way to (I am aware of my poor translation, but am not familiar with english terminology; however, I think you will understand.) "Reduce the following fraction:"

$\dfrac{x | x-1 |-x+1}{x^2-2|x|+1}$

besides doing numerous situations with absolute values (in this case, 4 situations);
in this case I'd make:
i.) $x - 1 \ge 0$, $\space x \le 0$

ii.) $x - 1 \le 0$, $\space x \ge 0$

iii.) $x - 1 \ge 0$, $\space x \ge 0$

iv.) $x - 1 \le 0$, $\space x \le 0$

Then solve each pair's cross-section of the two solutions...

Second thing I'm not certain about are the "$\ge$" and "$\le$" signs. Do I have to put on both equations in the pair the equal sign "$=$", or just "$>$" / "$<$". If it's not clear, do I have to make every like:

i.) $x - 1 \ge 0$, $\space x \le 0$

ii.) $x - 1 \le 0$, $\space x \ge 0$

or

i.) $x - 1 \ge 0$, $\space x < 0$

ii.) $x - 1 \le 0$, $\space x > 0$

Please explain when to put the "$\le$" and "$\ge$".

EDIT:
Basically, what I wanted to know is, if there is any kind of faster - automated - way of solving the question. The answers you provided matched my opinion and praxis. The thing here is, I know a faster way of solving these, it includes a, so called "table" (with 'null-points', something like RecklessReckoner advised), if someone is interested I'll make a photo of it. It makes solving these totally automated that you don't have to think about the actual problem and/or understand it, that's why I tried to solve these traditionally to understand it - behind the scenes. Once again, thanks for your answers.

Answer

It takes thinking. When you split into cases, generally you have something like $|x-a|$, which will be $x-a$ if $x \ge a$ and $a-x$ if $a \ge x$. If $x=a$ it doesn't matter which direction you use, so generally you use the sign with the one with the equals because you don't want to exclude a potential solution. The point is "do you know it isn't equal-otherwise include the equals. Then you need to check the solutions you find in the original equation because you may have introduced extraneous solutions.