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

Try to find the value of $x$.
(2003, National Junior High School Mathematics League)
Given:
$$
\begin{array}{l}
\frac{1}{b}=x-a, \frac{1}{d}=x-c, b=x-\frac{1}{c}, d=x-\frac{1}{a} . \\
\text { Then } (x-a)\left(x-\frac{1}{c}\right)=1, (x-c)\left(x-\frac{1}{a}\right)=1 .
\end{array}
$$

Thus, 
$$
c x^{2}-(a c+1) x+a-c=0,
$$
$$
a x^{2}-(a c+1) x+c-a=0 .
$$

Subtracting the two equations, we get
$$
(c-a) x^{2}=2(c-a) \text{. }
$$

Since $c \neq a$, it follows that $x^{2}=2, x= \pm \sqrt{2}$.