2. Zadatak

Prvo računamo umnožak

$$(a+b)\left(\frac1b-\frac1a\right)=\frac ab-\frac ba.$$

Tada dana jednakost prelazi u

$$\left[\left(\frac ab-\frac ba\right): \left(\frac ab-\frac ba\right)-\frac ab\right]: \left(\frac ab+1\right)=\frac1{2014},$$

odnosno

$$\frac{1-\frac ab}{\frac ab+1}=\frac1{2014} \qquad\text{ili oblik}\qquad \frac{b-a}{a+b}=\frac1{2014}.$$

Slijedi

$$\frac ab+1=2014-2014\cdot\frac ab,$$ $$2015\cdot\frac ab=2013,$$ $$\frac ab=\frac{2013}{2015}.$$