9. (16 points) Given four real numbers such that the product of any three of them plus the remaining one equals 10. Find these four real numbers.

保留源文本的换行和格式，直接输出翻译结果。
Let the four real numbers be $a, b, c, d$. Then,
$$
\left\{\begin{array}{l}
bcd + a = 10, \\
cda + b = 10, \\
dab + c = 10, \\
abc + d = 10.
\end{array}\right.
$$

Let $m = abcd$. Thus,
$$
\left\{\begin{array}{l}
a^2 - 10a + m = 0, \\
b^2 - 10b + m = 0, \\
c^2 - 10c + m = 0, \\
d^2 - 10d + m = 0.
\end{array}\right.
$$

Note that the equation $x^2 - 10x + m = 0$ has at most two distinct real roots.
Assume $m = 25$, then only the real number 5 satisfies the equation.
Thus, $a = b = c = d = 5$, and in this case,
$$
m = abcd = 5^4 \neq 25,
$$

which is a contradiction.
Therefore, $m \neq 25$, meaning the equation has two distinct real roots $x_1, x_2$.
It is easy to see that $m \neq 0$.
(1) $a = b = c = d$.
Assume $a = x_1$, then
$$
x_1^3 + x_1 = 10 \Rightarrow x_1 = 2.
$$
(2) $a, b, c, d$ are not all equal.

Assume $a \neq b$, then $\{a, b\} = \{x_1, x_2\}$.
From $x_1 x_2 = m = abcd = x_1 x_2 cd$
$\Rightarrow cd = 1$.
If $c \neq d$, then
$\{c, d\} = \{x_1, x_2\}, m = 1$
$\Rightarrow \{x_1, x_2\} = \{5 + 2\sqrt{6}, 5 - 2\sqrt{6}\}$;
If $c = d$, assume $c = x_1$, then
$x_1^2 = 1 \Rightarrow x_1 = 1$ or -1
$\Rightarrow x_2 = 9$ or 11.
In summary, the four real numbers are $2, 2, 2, 2$ or $5 + 2\sqrt{6}, 5 + 2\sqrt{6}, 5 - 2\sqrt{6}, 5 - 2\sqrt{6}$ or $9, 1, 1, 1$ or $11, -1, -1, -1$.