14・12 Solve the equation $[x] \mid x\}+x=2\{x\}+10$.
(Recommended by the Soviet Ministry of Education, 1991)
[Solution] Since $x=[x]+\{x\}$, the given equation can be transformed into
$$
\begin{array}{l}
{[x]\{x\}+[x]-\{x\}=10,} \\
([x]-1)(\{x\}+1)=9 .
\end{array}
$$

Since $[x]-1$ is an integer, then $\{x\}+1$ is a rational number.
Since $0 \leqslant\{x\}<1$, when $\{x\}+1>0$, we can assume $(n, k)=1$, then
$$
(n+k, k)=1 .
$$

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

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

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

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

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