Three, 15. The original equation simplifies to
$$k x^{2}-3 k x+2 x-1=0 \text {. }$$

When $k=0$, the original equation has a unique solution $x=\frac{1}{2}$.
When $k \neq 0$, for equation (1),
$$\Delta=(3 k-2)^{2}+4 k=5 k^{2}+4(k-1)^{2}>0 \text {. }$$

Therefore, there are always two distinct real roots. According to the problem, the original equation has only one solution, so one of the roots must be an extraneous root. From the original equation, the extraneous root can only be 0 or 1. Clearly, 0 is not a root of (1), so $x=1$ is a root of equation (1). Substituting into (1) gives $k=\frac{1}{2}$.

By Vieta's formulas, the root of the original equation is $-\frac{1}{k}=-2$.
Therefore, when $k=0$, the solution to the equation is $x=\frac{1}{2}$.
When $k=\frac{1}{2}$, the solution to the equation is $x=-2$.