One, (20 points) Given real numbers $a, b, c, d$ are all distinct and satisfy
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{a}=x \text {. }
$$

Find the value of $x$.
$$
\begin{array}{l}
d=x-\frac{1}{a}=\frac{a x-1}{a}, \\
c=x-\frac{1}{d}=x-\frac{a}{a x-1}=\frac{a x^{2}-x-a}{a x-1}, \\
b=x-\frac{1}{c}=x-\frac{a x-1}{a x^{2}-x-a} \\
=\frac{a x^{3}-x^{2}-2 a x+1}{a x^{2}-x-a}, \\
a+\frac{a x^{2}-x-2 a}{a x^{3}-x^{2}-2 a x+1}=x .
\end{array}
$$

Therefore, $a x^{4}-a^{2} x^{3}-x^{3}-2 a x^{2}+2 a^{2} x+2 x=0$.
Factoring gives
$$
x\left(x^{2}-2\right)\left(a x-a^{2}-1\right)=0 .
$$

Thus, $x=0$ or $x^{2}=2$ or $a x-a^{2}-1=0$.
If $x=0$, then from $a+\frac{1}{b}=b+\frac{1}{c}=x=0$ we have $b=c$, which contradicts the given conditions;

If $a x-a^{2}-1=0$, then $x=a+\frac{1}{a}$, but $x=a+\frac{1}{b}$, thus, $a=b$, which contradicts the given conditions.
Hence, $x^{2}=2$, i.e., $x=\sqrt{2}$ or $-\sqrt{2}$.