6. If the equation $\frac{2 k}{x-1}-\frac{x}{x^{2}-x}=\frac{k x+1}{x}$ has only 1 solution, then $k=$
6. $k=0$ or $\frac{1}{2}$.

The original equation, after clearing the denominator and rearranging, becomes
$$
k x^{2}-(3 k-2) x-1=0 \text {. }
$$

If $k=0$, the only solution to equation (1) is $x=\frac{1}{2}$, which is also the only solution to the original equation.
If $k \neq 0$, then
$$
\Delta=(3 k-2)^{2}+4 k=9\left(k-\frac{4}{9}\right)^{2}+\frac{20}{9}>0,
$$

which means equation (1) always has two real roots, one of which is an extraneous root of the original equation. The extraneous root of the original equation can only be 0 or 1, but $x=0$ is not a root of equation (1), so the extraneous root is $x=1$. Substituting $x=1$ into equation (1), we get
$$
k-(3 k-2)-1=0, k=\frac{1}{2} \text {. }
$$

In summary, $k=0$ or $\frac{1}{2}$.