Three. (25 points) Given real numbers $a, b, c, d$ are all distinct, and
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{d}=d+\frac{1}{a}=x \text {. }
$$

Find the value of $x$.
Three, Solution 1: From the given, we have
$$
\begin{array}{l}
a+\frac{1}{b}=x, \\
b+\frac{1}{c}=x, \\
c+\frac{1}{d}=x, \\
d+\frac{1}{a}=x .
\end{array}
$$

From equation (1), we solve for
$$
b=\frac{1}{x-a} \text {. }
$$

Substituting equation (5) into equation (2) gives
$$
c=\frac{x-a}{x^{2}-a x-1} \text {. }
$$

Substituting equation (6) into equation (3) gives
$$
\frac{x-a}{x^{2}-a x-1}+\frac{1}{d}=x \text {, }
$$

which simplifies to $d x^{3}-(a d+1) x^{2}-(2 d-a) x+a d+1=0$.
From equation (4), we get $a d+1=a x$, substituting this into equation (7) gives
$$
(d-a)\left(x^{3}-2 x\right)=0 \text {. }
$$

Given that $d-a \neq 0$, hence $x^{3}-2 x=0$.
If $x=0$, then from equation (6) we get $a=c$, which is a contradiction. Therefore,
$$
x^{2}=2, x= \pm \sqrt{2} \text {. }
$$