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.
 A: *

*$$\frac{1}{xy} = 2005(\frac{1}{x}+\frac{1}{y}) \iff \frac{1}{xy}=\frac{2005(x+y)}{xy}$$
which follows immediately from the condition




*$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2=$$$$= \frac{y}{y-x}-({1}-\frac{x}{y})-\frac{x}{y-x}-(-1+\frac{y}{x})+\frac{y}{x} - \frac{x}{y}+2=$$$$=\frac{y-x}{y-x}+2=3$$
Explanation:

*

*First divide the fraction into two fractions (like $\frac{y-x}{y}=1-\frac{x}{y}$)

*Cancel out the opposite terms

A: 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:
$$\begin{align}
{y\over y-x}-{y-x\over y}-{x\over y-x}-{y-x\over x}+{y\over x}-{x\over y}+2
&=\left({y\over y-x}-{x\over y-x} \right)-\left({y-x\over y}+{x\over y} \right)+\left({y\over x}-{y-x\over x} \right)+2\\
&=\left(y-x\over y-x\right)-\left(y\over y\right)+\left(x\over x\right)+2\\
&=1-1+1+2\\
&=3
\end{align}$$
A: For the second part of your question,
For $y \neq x$,
$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l = \left(\frac{y}{y-x} - \frac{x}{y-x}\right) - \frac{y-x}{y} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2 $$
$$\implies l= -(y-x)\left(\frac{1}{y} + \frac{1}{x}\right) + \frac{(y+x)(y-x)}{xy} + 3$$
$$\implies l = (y-x)\left(\frac{y+x}{xy} - \frac1y - \frac1x\right) + 3$$
$$\implies l =3$$
A: 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 $3$.
