$14 \cdot 24$ Solve the equation $x+\frac{92}{x}=[x]+\frac{92}{[x]}$,
(18th All-Russian Mathematical Olympiad, 1992)
【Sol】From the equation, we know $x \neq 0, \quad[x] \neq 0$, and
$$
\begin{array}{l}
(x-[x])+\left(\frac{92}{x}-\frac{92}{[x]}\right)=0, \\
(x-[x])\left(1-\frac{92}{x[x]}\right)=0 .
\end{array}
$$

Thus, $x=[x]$ or $\frac{92}{x[x]}=1$. When $x=[x]$, $x=n$, where $n$ is any non-zero integer.
When $\frac{92}{x[x]}=1$,
$$
x[x]=92 \text{. }
$$

Let $x=[x]+a,[x]=n, \quad n \in \mathbb{Z}$, and $0 \leqslant a < 1$. When $n > 0$, $n^{2} \leqslant 92 < n(n+1)$.

In this case, the integer solution is $n=-10$. Thus,
$$
\begin{array}{c}
(-10)(-10+a)=92, \\
a=0.8 .
\end{array}
$$

Therefore,
$$
x=-10+0.8=-9.2 \text{. }
$$

Hence, the solutions to this problem are
$x=n$, where $n$ is a non-zero integer, or $x=-9.2$.