13. For the equation $k x^{2}-(k-1) x+1=0$ with respect to $x$, if it has rational roots, find the integer value of $k$.
Three, 13. (1) When $k=0$, $x=-1$, the equation has a rational root.
(2) When $k \neq 0$, since the equation has a rational root, if $k$ is an integer, then $\Delta=(k-1)^{2}-4 k=k^{2}-6 k+1$ must be a perfect square, i.e., there exists a non-negative integer $m$, such that
$$
k^{2}-6 k+1=m^{2} .
$$

Completing the square, we get $(k-3)^{2}-m^{2}=8$.
That is, $(k-3+m)(k-3-m)=8$.
Since $k-3+m$ and $k-3-m$ are integers with the same parity, and their product is 8, they must both be even.
Also, $k-3+m \geqslant k-3-m$, so we have
$$
\left\{\begin{array} { l } 
{ k - 3 + m = 4 } \\
{ k - 3 - m = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
k-3+m=-2 . \\
k-3-m=-4 .
\end{array}\right.\right.
$$

Solving, we get $k=6$ or $k=0$ (discard).
Combining (1) and (2), we know that when the equation $k x^{2}-(k-1) x+1=0$ has a rational root, $k=6$ or $k=0$.