10. Solve the equation:

$$
\frac{8}{\{x\}}=\frac{9}{x}+\frac{10}{[x]}
$$

where $[x]$ is the greatest integer less than or equal to $x$, and $\{x\}=x-[x]$.
Answer: $\frac{3}{2}$.

Solution: Note that $x$ cannot be negative, because in such a case the right-hand side would be negative. Moreover, $x$ cannot be an integer, because the fractional part would be 0, and it cannot belong to the interval $[0,1)$, because the integer part would be 0. Let's consider the possible cases:

- If $[x]=1$, then $\frac{8}{\{x\}}=\frac{9}{1+\{x\}}+10$. Then $\{x\}=\frac{1}{2}$, hence $x=\frac{3}{2}$;
- If $[x]=2$ then $\frac{8}{\{x\}}=\frac{9}{2+\{x\}}+5$, from which we get $\{x\}=1$, which is impossible.
- If $[x] \geqslant 3$, then $\frac{9}{x}+\frac{10}{[x]}<3+\frac{10}{3}<8<\frac{8}{\{x\}}$.