How to separate denominator? I am sorry,
it's probably very basic question, but I have great hole in head and can't figure it out. I need to separate denominator, like that:
$ \frac { 1 } { n _ { 0 } + n _ { 1 } + n _ { 2 } + \ldots } = \frac { A } { n _ { 0 } } + \frac { B } { n _ { 1 } } + \frac { C } { n _ { 2 } } + \dots $
Please could anyone give some hint. For any help thanks in advance.
 A: There are too many degrees of freedom. This means there are many (infinite even) ways to do this.
For instance, consider the case with two variables:
$$ \frac{1}{a+b} = \frac{c}{a} + \frac{d}{b} $$
Notice that you have one equation with two unknown variables! 
This means we can simply set $c$ to be some constant, e.g. $1$, and then solve for $d$, getting:
$$ d = b\left( \frac{1}{a+b} - \frac{c}{a} \right)  = b\left( \frac{1}{a+b} - \frac{1}{a} \right) $$
for $c=1$. Of course you could choose anything for $c$.
Let's do it in general. Let:
$$ S = \left[ \sum_{i=1}^m x_i \right]^{-1} $$
Then we want to find $\{a_i\}_{i=1}^m$ such that:
$$ S = \sum_{i=1}^m \frac{a_i}{x_i} $$
Well, since there is one equation, we have $m-1$ degrees of freedom. Let us fix all the parameters except $a_j$. Essentially we shall freely choose values for $\{a_i\}_{i\ne j}$, at the cost of $a_j$ not being free (i.e. $a_j$ is constrained by the values we choose for $a_i$. Solving for $a_j$ we get:
$$
S =\frac{a_j}{x_j} + \sum_{i\ne j} \frac{a_i}{x_i} \;\;\;\implies\;\;\;
a_j = x_j\left( S - \sum_{i\ne j} \frac{a_i}{x_i} \right)
$$
Cool. So now we can choose some values for the $a_i$'s, and this will fix the value of $a_j$. 

For example, let's choose $a_i=x_i\;\forall\;i\ne j$. Then we get:
$$ a_k = \begin{cases}
x_k & k\ne j\\
x_j(S^{-1}-m+1) & k=j
\end{cases} $$

As another example, consider: $ a_i = x_iS/m $ for all $i\ne j$. Then what is $a_j$? Well: $$ a_j=x_j\left( S - \sum_{i\ne j}\frac{S}{m} \right)=x_j\frac{S}{m} $$
So we can say $a_k = x_k S/m$ for all $i\in[1,m]$ for this case.
