12. (15th American Mathematical Invitation Contest) Given that $a, b, c, d$ are non-zero real numbers, $f(x)=\frac{a x+b}{c x+d} (x \in \mathbf{R})$, and $f(19)=19, f(97)=97$. If for any real number $x$, when $x \neq-\frac{d}{c}$, $f[f(x)]=x$, find the only number that is not in the range of $f(x)$.
12. By the given condition, for any real number $x \neq-\frac{d}{c}$, we have $f[f(x)]=x$, so $\frac{a \cdot \frac{a x+b}{c x+d}+b}{c \cdot \frac{a x+b}{c x+d}+d}=x$, simplifying to $(a+d) c x^{2}+$ $\left(d^{2}-a^{2}\right) x-b(a+d)=0$. Since this equation holds for $x \neq-\frac{d}{c}$, we have $a+d=0$, and $d^{2}-a^{2}=0$, so $d=-a$.

Also, $f(19)=19, f(97)=97$, meaning 19 and 97 are the roots of the equation $\frac{a x+b}{c x+d}=x$, or $c x^{2}+(d-a) x-b=0$. By Vieta's formulas, we get $\frac{a-d}{c}=116, -\frac{b}{c}=-1843$, combining with $d=-a$, we have $a=58 c, b=-1843 c, d=-58 c$, thus $f(x)=\frac{58 x-1843}{x-58}=58+\frac{1521}{x-58}$. Therefore, $f(x)$ cannot take the value 58, which is the only number outside the range of $f(x)$.