For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$

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*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$


*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
For the first question, I tried to work it out with algebra; I solved for x through the equation given, then multiplied it by y and I got the value of $\frac{1}{xy} = 2005\left(\frac{1}{y-2005y^2}\right) $. Then I tried proving that $\frac{1}{y-2005y^2} =\frac{1}{x} + \frac{1}{y} $ but I failed at this.

According to the problem, $x, y$ $\in$ $\mathbb{R}$ and $x \neq y$.
$\bullet$ For the first part,
(I) From the given, we have that
\begin{align*}
&2005(x + y) = 1\\
\implies & 2005 \cdot \frac{(x + y)}{xy} = \frac{1}{xy}\\
\implies & 2005 \cdot \bigg( \frac{1}{x} + \frac{1}{y} \bigg) = \frac{1}{xy}
\end{align*}
Hence, done!
$\bullet$ For the second part,
(II)  According to question,
\begin{align*}
l =& ~\frac{y}{y-x} - \frac{y - x}{y} - \frac{x}{y - x} - \frac{y - x}{x} + \frac{y}{x} - \frac{x}{y} + 2\\
= & ~ \bigg[ \frac{y}{y-x} - \frac{x}{y - x} \bigg] - \bigg[ \frac{y - x}{y} + \frac{x}{y} \bigg] + \bigg[ \frac{y}{x} - \frac{y - x}{x} \bigg] + 2\\
= & ~ 1 - 1 + 1 + 2\\
= &~ 3
\end{align*}
Hence, the value of $l$ is $3$.
