-、(20 points) Let the real number $k$ satisfy $0<k<1$. Solve the fractional equation about $x$
$$
\frac{2 k}{x-1}-\frac{1}{x^{2}-x}=\frac{k+1}{x} \text {. }
$$
$$
\begin{array}{l}
\frac{2 k x}{(x-1) x}-\frac{1}{(x-1) x}=\frac{(k+1)(x-1)}{(x-1) x} \\
\Rightarrow 2 k x-1=(k+1)(x-1) \\
\Rightarrow(k-1) x=-k . \\
\text { Also, } 0<k<1 \text {, so } x=-\frac{k}{k-1} .
\end{array}
$$

Upon inspection, when $k=\frac{1}{2}$, $x=1$ is an extraneous root, and the original equation has no solution;
(Continued on page 48)
When $0<k<1$, and $k \neq \frac{1}{2}$, the solution to the original equation is
$$
x=-\frac{k}{k-1} .
$$