5. Represent the number $\frac{3}{7}$ as the sum of several different common fractions, the numerators of which are equal to one.

#
# Solution

We use the fact that

$$
\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}
$$

Then

$$
\begin{aligned}
& \frac{3}{7}=\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{1}{7}+\frac{1}{8}+\frac{1}{56}+\frac{1}{8}+\frac{1}{56}= \\
& =\frac{1}{7}+\frac{1}{8}+\frac{1}{56}+\frac{1}{9}+\frac{1}{72}+\frac{1}{57}+\frac{1}{56 \cdot 57}
\end{aligned}
$$

Other variations are possible.

Any other correct representation was also accepted as an answer.