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.
 A: 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.
A: I generally tell students to approach functions containing expressions in absolute value brackets this way.  The expression $ \ \vert u(x) \vert \ $ in brackets introduces "special points" at the values of $ \ x \ $ where $ \  u(x) = 0  \ $.  To either side of these points will be intervals where the expression is read as $ \ u(x) \ $ if  $ \ u(x) > 0 \ $  and $ \ -u(x) \ $ if  $ \ u(x) < 0 \ $  .  This reduces the matter of interpretation to one of reading the behavior of the function in each interval, rather than a "logic puzzle" requiring consideration of cases based on a collection of "binary switches", so to speak.
As you and the commenters have noted, there are two "special values" at $ \ x = 0 \ $ and $ \ x = 1 \ $ , so, all told, there are five possible "readings" of the rational function in question:
for $ \ x < 0 \ , $  where $ \ \vert x \vert = -x \ $ and $ \ \vert x-1 \vert = 1-x \ ; $
for $ \ x = 0 \ , $  where $ \ \vert x \vert = 0 \ $ and $ \ \vert x-1 \vert = 1 \ ; $
for $ \ 0 < x < 1 \ , $  where $ \ \vert x \vert = x \ $ and $ \ \vert x-1 \vert = 1-x \ $ ;
for $ \ x = 1 \ , $  where $ \ \vert x \vert = 1 \ $ and $ \ \vert x-1 \vert = 0 \ $ ; and
for $ \ x >  1 \ , $  where $ \ \vert x \vert = x \ $ and $ \ \vert x-1 \vert = x - 1 \ $ .
(This last produces an especially simple result...)
ADDENDUM: That being said, this particular rational function has an additional complication at $ \ x = -1 \ , $ owing to the form of the denominator function.  This, it seems, is not evident from looking at the absolute value expressions alone... 
