Task B-1.2. If $\left[\left(\frac{a}{b}-\frac{b}{a}\right):\left((a+b)\left(\frac{1}{b}-\frac{1}{a}\right)\right)-\frac{a}{b}\right]:\left(\frac{a}{b}+1\right)=\frac{1}{2014}$, what is $\frac{a}{b}$?
## Solution.

First, we calculate the product

$$
(a+b)\left(\frac{1}{b}-\frac{1}{a}\right)=\frac{a}{b}-\frac{b}{a}
$$

Then the given equation becomes

$$
\left[\left(\frac{a}{b}-\frac{b}{a}\right):\left(\frac{a}{b}-\frac{b}{a}\right)-\frac{a}{b}\right]:\left(\frac{a}{b}+1\right)=\frac{1}{2014}
$$

or equivalently

$$
\frac{1-\frac{a}{b}}{\frac{a}{b}+1}=\frac{1}{2014} \quad\left(\text { or the form } \frac{b-a}{a+b}=\frac{1}{2014}\right)
$$

It follows that

$$
\begin{aligned}
\frac{a}{b}+1 & =2014-2014 \cdot \frac{a}{b} \\
2015 \cdot \frac{a}{b} & =2013 \\
\frac{a}{b} & =\frac{2013}{2015}
\end{aligned}
$$