One. (20 points) Equations
$$
x^{2}+a x+b=0 \text { and } x^{2}+b x+a=0
$$

have a common root, and let the other two roots be $x_{1} 、 x_{2}$; Equations
$$
x^{2}-c x+d=0 \text { and } x^{2}-d x+c=0
$$

have a common root, and let the other two roots be $x_{3} 、 x_{4}$. Find the range of $x_{1} x_{2} x_{3} x_{4}$ $(a, b<0, a \neq b, c, d<0, c \neq d)$.
Let the common root of the first two equations be $x_{0}$. Then
$$
\left\{\begin{array}{l}
x_{0}^{2} + a x_{0} + b = 0, \\
x_{0}^{2} + b x_{0} + a = 0 .
\end{array}\right.
$$

Subtracting the two equations gives $(a-b) x_{0} = a-b \Rightarrow x_{0} = 1$.
By Vieta's formulas, the other two roots are
$$
x_{1} = \frac{b}{1} = b, x_{2} = \frac{a}{1} = a \text{, and } 1 + a + b = 0 \text{. }
$$

Then $0 < x_{1} x_{2} = a b = a(-1-a)$
$$
= -\left(a + \frac{1}{2}\right)^{2} + \frac{1}{4} \leqslant \frac{1}{4} \text{. }
$$

Similarly, $0 < x_{3} x_{4} = (-c)(-d) = c d$
$$
= c(-1-c) = -\left(c + \frac{1}{2}\right)^{2} + \frac{1}{4} \leqslant \frac{1}{4} \text{. }
$$

Therefore, $0 < x_{1} x_{2} x_{3} x_{4} \leqslant \frac{1}{16}$.
The maximum value $\frac{1}{16}$ is achieved if and only if $a = b = c = d = -\frac{1}{2}$. Since $a \neq b, c \neq d$, the equality cannot be achieved.
Hence, $0 < x_{1} x_{2} x_{3} x_{4} < \frac{1}{16}$.