4. The fractional equation about $x$ $\frac{1}{x+2}-\frac{k}{x-2}=1$ $-\frac{4 x}{x^{2}-4}$ has two real roots. Then, $k$ should satisfy ( ).
(A) $k^{2}-18 k+33>0$
(B) $k^{2}-18 k+33>0$ and $k \neq 2$
(C) $k \neq 2$
(D) None of the above answers is correct
4. (B).

Eliminating the denominator and rearranging gives
$$
x^{2}+(k-5) x+(2 k-2)=0 \text {. }
$$

According to the problem, $k$ should satisfy
$$
\left\{\begin{array}{l}
\Delta=(k-5)^{2}-4(2 k-2)=k^{2}-18 k+33>0, \\
2^{2}+2(k-5)+2 k-2 \neq 0, \\
(-2)^{2}-2(k-5)+2 k-2 \neq 0 .
\end{array}\right.
$$

That is, $k^{2}-18 k+33>0$ and $k \neq 2$.