3. If the equation about $x$
$$
\frac{2 k}{x+2}-\frac{x-1}{x^{2}+2 x}=\frac{k x+1}{x}
$$

has only one solution (equal solutions are counted as one), then the value of $k$ is $\qquad$
3.0 or 1 or $\frac{3}{4}$.
The original fractional equation can be transformed into
$$
k x^{2}+2 x+1=0 \text {. }
$$
(1) When $k=0$, $x=-\frac{1}{2}$ is a solution to the fractional equation.
(2) When $k \neq 0, \Delta=2^{2}-4 k=0$, $k=1$, $x=-1$ is a solution to the fractional equation.
(3) When $k \neq 0, \Delta>0$, one of the roots of equation (1) is an extraneous root of the fractional equation, and the other is a root of the fractional equation. When $x=-2$ is an extraneous root of the fractional equation, substituting it into equation (1) yields
$$
k(-2)^{2}+2 \times(-2)+1=0 \Rightarrow k=\frac{3}{4} .
$$

At this point, the other root of equation (1), $x=-\frac{2}{3}$, is a solution to the fractional equation; when $x=0$ is an extraneous root of the fractional equation, substituting it into equation (1) yields $k \times 0^{2}+2 \times 0+1=0$, which is impossible.
Therefore, the value of $k$ is 0 or 1 or $\frac{3}{4}$.