Find all triplets of three strictly positive integers such that:

$$
x+\frac{1}{y+\frac{1}{z}}=\frac{10}{7}
$$
Note that $y+\frac{1}{z}>1$ so $x \leqslant x+\frac{1}{y+\frac{1}{z}}<x+1$. By taking the integer part of both sides of the equation, we find that $x=1$, hence $\frac{1}{y+\frac{1}{z}}=\frac{3}{7}$. From this, we deduce that $y+\frac{1}{z}=\frac{7}{3}$. By taking the integer part, we get $y=2$ and then $z=3$.