The correct equation for calculating parallel resistance is: \$ \frac{1}{R_{p}}= \frac{1}{R_{1}} + \frac{1}{R_{2}} + ...\frac{1}{R_{n}}\$, for however many different resistances you have in parallel.
The alternate form of \$ \frac{1}{x} + \frac{1}{y} \$ is \$ \frac{x+y}{xy} \$. This is true for any numbers, try it yourself if you're not sure.
So in the limited case where there are only 2 resistors in parallel, we can alternatively say \$ \frac{1}{R_{p}}= \frac{R_{1}+R_{2}}{R_{1}R_{2}}\$. Taking the inverse of both sides results in \$ R_{p}= \frac{R_{1}R_{2}}{R_{1}+R_{2}}\$
Your statement that \$ \frac{1}{R_{p}}= \frac{R_{2}}{R_{1}R_{2}} + \frac{R_{1}}{R_{1}R_{2}}\$, while technically correct, is simply an unsimplified form of what was already stated. As the \$ R_{2}\$ terms cancel in the fraction \$ \frac{R_{2}}{R_{1}R_{2}}\$, leaving simply \$ \frac{1}{R_{1}} \$ and vise versa for the other term.
All you're doing is making it more complicated that necessary. As to why your equations don't seem to work out, it is because you are not evaluating it correctly.
The inverse of \$ \left ( \frac{R_{2}}{R_{1}R_{2}} + \frac{R_{1}}{R_{1}R_{2}} \right ) \$ is not equal to \$ \frac{R_{1}R_{2}}{R_{2}} + \frac{R_{1}R_{2}}{R_{1}}\$. Exponentiation cannot be distributed over multiple terms unless those terms are multiplicative. You are distributing the exponentiation as if the two terms were being multiplied, rather than added, and that is incorrect. You must invert the entire expression, yielding (and this is why this is needlessly overcomplicating things):
\$ R_{p}= \frac{1}{\frac{R_{2}}{R_{1}R_{2}} + \frac{R_{1}}{R_{1}R_{2}}} \$. This is correct, if complicated.
R1/R1R2equals? \$\endgroup\$ – Andy aka Aug 25 '20 at 7:44