How to remove parentheses from $x/(y-z)$ To remove parentheses from $x(y-z)$ I reword it to $xy-xz$.
How do I remove parenthesis from $x/(y-z)$?
 A: N.B. I will assume that we are only dealing with real numbers here.
I can very well imagine what you mean to ask, eventhough "removing parntheses form $x/(y-z)$" has no conventional meaning.

What you are dealing with is called the distributive property. We say that multiplication ($*$) is left-distributive over addition ($+$), because $$a*(b+c)=ab+bc\tag L\label l$$
always holds. Likewise, we have right-distributivity because $$(a+b)*c=ac+bc\tag R\label r$$ always holds.  We say that addition is not left-distributive over  multiplication, because $$a+(b*c)=a+b*a+c$$ does not always hold.
Notice how the word "because" is emphasised. The distibutive property is not something we deliberately construct. It either holds, or it doesn't. This also means that the way to see if a property is (left-, or right-)distributive, is to simply check if $\eqref l$ and/or $\eqref r$ always hold(s).

Now for your case, division ($/$) is not left-distributive over substraction ($-$). Look for instance at $$12/(4-6)\neq 12/ 4-12/ 6.$$
However, it is right-distributive: $$(a- b)/ c=a/ c-b/ c,$$ always.

So to sum up:

*

*removing the parentheses from $x/(y-z)$ has no clear meaning

*we cannot use the distributive property here to find an expression, which equals $x/(y-z)$ and contains no parentheses.

As far as I can see, the only thing we could say is that $x/(y-z)=\frac{x}{y-z}$, which is just another notational convention for writing division (so it wouldn't really count as algebraic manipulation).
A: How about $$x/(y-z) = \frac{x}{y} \frac{1}{(1-\frac{z}{y})} = \frac{x}{y} (1 + \frac{z}{y} + \frac{z^2}{y^2} + \cdots) = \sum\limits_{i=0}^{\infty} \frac{xz^i}{y^{i+1}}$$
