<s>[INST] 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 $2005(x+y) = 1$

*

*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.
 [/INST] For part 1, if $xy\not=0$, then
$$\begin{align}
1=2005(x+y)\implies{1\over xy}&=2005(x+y){1\over xy}\\
&=2005\left({x\over xy}+{y\over xy}\right)\\
&=2005\left({1\over y}+{1\over x}\right)\\
&=2005\left({1\over x}+{1\over y}\right)
\end{align}$$
For part 2, regroup the terms with denominators $y-x$, $y$, and $x$ and simplify:
$$\begin{align}
{y\over y-x}-{y-x\over y}-{x\over y-x}-{y-x\over x}+{y\over x}-{x\over y}+2
&=\left({y\over y-x}-{x\over y-x} \right)-\left({y-x\over y}+{x\over y} \right)+\left({y\over x}-{y-x\over x} \right)+2\\
&=\left(y-x\over y-x\right)-\left(y\over y\right)+\left(x\over x\right)+2\\
&=1-1+1+2\\
&=3
\end{align}$$
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