1. (51st Czech and Slovak Mathematical Olympiad (Final) Question) Find $a, b$ in the real numbers such that the equation $\frac{a x^{2}-24 x+b}{x^{2}-1}=x$
has two roots, and the sum of these roots equals 12, where a repeated root counts as one root.
1. Transform and simplify the original equation to get
$$
x^{3}-a x^{2}+23 x-b=0 \text {. }
$$

Since a cubic equation has either one root or three roots in the real number range, these three roots are either one root being 1 or -1, and the other two roots are distinct; or no root is $\pm 1$, but there is a double root. Since the sum of these two roots is 12, the three roots can only be one of the following two cases:
$(-1, s, 12-s) 、(1, s, 12-s),(s, s, 12-s)$, and $s \notin\{-1,1,6,11,13\}$.
If $\left(x_{1}, x_{2}, x_{3}\right)=(-1, s, 12-s)$, by Vieta's theorem we get
$$
a=11, s^{2}-12 s+35=0, b=-s(12-s) \text {. }
$$

Therefore, $s=5$ or $7, b=-35$.
Thus, $(a, b)=(11,-35)$.
If $\left(x_{1}, x_{2}, x_{3}\right)=(s, s, 12-s)$, similarly we get
$$
a=s+12, s^{2}-24 s+23=0, b=s^{2}(12-s) .
$$

Therefore, $s=1$ or 23.
Discarding $s=1$, we get $a=35, b=-5819$.
Thus, $(a, b)=(35,-5819)$.
In summary, there are two pairs of $(a, b)$ that satisfy the conditions, which are $(11,-35),(35,-5819)$.