4・147 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{x_{2} x_{3} x_{4} \cdots x_{n}}{x_{1}}=a_{1} \\
\frac{x_{1} x_{3} x_{4} \cdots x_{n}}{x_{2}}=a_{2} \\
\cdots \cdots \cdots \cdots \\
\frac{x_{1} x_{2} x_{3} \cdots x_{n-1}}{x_{n}}=a_{n}
\end{array}\right.$$
[Solution] First, consider the case where all $a_{i}>0$:
(1) If all $x_{k}(k=1,2, \cdots, n)$ are greater than 0.
Multiply the left and right sides of all equations, we get

Thus,
$$\left(x_{1} x_{2} x_{3} \cdots x_{n}\right)^{n-2}=a_{1} a_{2} a_{3} \cdots a_{n}$$

Rewrite the $k$-th equation as
$$a_{k} x_{k}^{2}=x_{1} x_{2} x_{3} \cdots x_{n}$$

By (1) and all $x_{i}>0$, we have
$$x_{k}=\sqrt{\frac{\sqrt[n-2]{a_{1} a_{2} \cdots a_{n}}}{a_{k}}}, k=1,2, \cdots, n$$
(2) If not all $x_{k}(k=1,2, \cdots, n)$ are greater than 0.
Obviously, changing the sign of any even number of $x_{k}$ in the solution of (1) still gives a solution to the original system of equations.
Now consider the case where all $a_{i}<0$:
Transform the original system of equations into
$$\left\{\begin{array}{l}
\frac{x_{2} x_{3} x_{4} \cdots x_{n}}{\left(-x_{1}\right)}=-a_{1} \\
\frac{\left(-x_{1}\right) x_{3} x_{4} \cdots x_{n}}{x_{2}}=-a_{2} \\
\cdots \cdots \cdots \cdots \\
\frac{\left(-x_{1}\right) \cdot x_{2} \cdot x_{3} \cdots x_{n-1}}{x_{n}}=-a_{n}
\end{array}\right.$$

Thus, by (1) and (2), its solution is
$$\begin{array}{l}
x_{1}=-\sqrt{\frac{\sqrt[n-2]{(-1)^{n} a_{1} a_{2} \cdots a_{n}}}{-a_{1}}} \\
x_{k}=\sqrt{\frac{\sqrt[n-2]{(-1)^{n} a_{1} a_{2} \cdots a_{n}}}{-a_{k}}}(k=2,3, \cdots, n)
\end{array}$$

Or change the sign of any even number of values among $x_{1}, x_{2}, \cdots, x_{n}$.
Note that $a_{1}, a_{2}, \cdots, a_{n}$ have the same sign as $x_{1} x_{2} x_{3} \cdots x_{n}$, so all $a_{i}$ have the same sign and cannot be zero. In other words, our discussion of the solutions to the system of equations is now complete.