9. (1991 USSR Ministry of Education Test) Solve the equation $[x]\{x\}+x=2\{x\}+10$.

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9. Since $x=[x]+\{x\}$, the given equation can be transformed into
$[x]\{x\}+[x]-\{x\}=10$,
$([x]-1)(\{x\}+1)=9$.
Since $[x]-1$ is an integer, then $\{x\}+1$ is a rational number.
Since $0 \leqslant\{x\}<1$, when $0 \leqslant\{x\}0$, we can assume $(n, k)=1$, at this time
$$
(n+k, k)=1 \text {. }
$$

Thus, 9 should be a multiple of $n+k$, so
$n+k=3$ or 9.
(1) When $n+k=3$, from $0 \leqslant n<k$ we get
$$
\begin{array}{l}
n=1, k=2, \\
\{x\}=\frac{1}{2},[x]=7 .
\end{array}
$$
(2) When $n+k=9$,

If $n=1, k=8$, then
$$
\{x\}=\frac{1}{8},[x]=9 \text {. }
$$

If $n=2, k=7$, then
$$
\{x\}=\frac{2}{7},[x]=8 \text {. }
$$

If $n=4, k=5$, then
$$
\{x\}=\frac{4}{5},[x]=6 \text {. }
$$

Thus, the given equation has 5 solutions:
$$
x=10,7 \frac{1}{2}, 9 \frac{1}{8}, 8 \frac{2}{7}, 6 \frac{4}{5} .
$$