$2 \cdot 118$ If the equation with respect to $x$
$$
\frac{1}{x^{2}-x}+\frac{k-5}{x^{2}+x}=\frac{k-1}{x^{2}-1}
$$

has an extraneous root $x=1$, then the value of $k$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 6
(China Beijing Junior High School Mathematics Competition, 1985)
[Solution] From $\frac{1}{x^{2}-x}+\frac{k-5}{x^{2}+x}=\frac{k-1}{x^{2}-1}$,
which is $\frac{1}{x(x-1)}+\frac{k-5}{x(x+1)}=\frac{k-1}{(x-1)(x+1)}$.
Multiplying both sides by $x(x-1)(x+1)$, we get
$$
(x+1)+(k-5)(x-1)=(k-1) x
$$

At this point, $x=1$ (extraneous root) should satisfy $*$, substituting into $\circledast$ we get $\quad k=3$. Therefore, the answer is $(C)$.