$$\begin{array}{l} \frac{2 k x}{(x-1) x}-\frac{1}{(x-1) x}=\frac{(k+1)(x-1)}{(x-1) x} \\ \Rightarrow 2 k x-1=(k+1)(x-1) \\ \Rightarrow(k-1) x=-k . \\ \text { Also, } 0<k<1 \text {, so } x=-\frac{k}{k-1} . \end{array}$$

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
$$x=-\frac{k}{k-1} .$$