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.

For the second part of your question,
For $y \neq x$,
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l = \left(\frac{y}{y-x} - \frac{x}{y-x}\right) - \frac{y-x}{y} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l= -(y-x)\left(\frac{1}{y} + \frac{1}{x}\right) + \frac{(y+x)(y-x)}{xy} + 3$$
$$\implies l = (y-x)\left(\frac{y+x}{xy} - \frac1y - \frac1x\right) + 3$$
$$\implies l =3$$
