Functions validity. Why does writing a function differently make it valid for a originally invalid input? 
$e.g:$
$$f(x) = \frac{1} {(\frac1x+2)(\frac1x-3)} \implies x≠0$$
Which  may alternatively be written as:
$$f(x) =\frac{x^2}{(1+2x)(1-3x)}$$
Which is valid for $x=0$?
Both graphically represent the same function. 
Thanks. 
 A: They aren't quite the same function.  They're the same function on all values where both are defined.  The domain of the first is $\mathbb{R}\setminus \{0, -\frac{1}{2}, \frac{1}{3}\}$.  The domain of the second is $\mathbb{R}\setminus \{-\frac{1}{2}, \frac{1}{3}\}$.  Those aren't the same domain, so the two functions aren't the same.  For any value in both of those domains (that is, any value but $x = 0$), they give the same value.  But the second has an extra point.
A: Like others have pointed out, the graph of both is not the same. There is a nontrivial singularity for the first function that is not present for the second.
A simpler version of your question seems to be:

Are the functions $f(x)=x$ and $f(x) = \frac{x^2}{x}$ the same?

The answer is, yes, except at the point $x=0.$ They are different because by default, mathematicians restrict the domain such that there can never be an input that requires division by zero. There isn't some mystical decree that states this should be the case, but it is generally assumed when a function has division by something that can be zero.
A: let $$g(x)=\frac{x^2}{(1+2x)(1-3x)}$$ and $$f(x) = \frac{1} {(\frac1x+2)(\frac1x-3)}$$
we have 
$D_f=R-\{0,\frac{1}{3},\frac{-1}{2}\}$ and $D_g=R-\{\frac{1}{3},\frac{-1}{2}\}$, thus $g\ne f$
