$4 \cdot 261$ Find all integers $k$, such that there exists an integer $x$, satisfying the equation
$$
\sqrt{39-6 \sqrt{12}}+\sqrt{k x(k x+\sqrt{12})+3}=2 k .
$$
[Solution] It is easy to know that
$$
\begin{array}{l}
39-6 \sqrt{12}=(6-\sqrt{3})^{2}, \\
k x(k x+\sqrt{12})+3=(k x+\sqrt{3})^{2} .
\end{array}
$$

Therefore, the original equation can be transformed into
$$
6-\sqrt{3}+|k x+\sqrt{3}|=2 k,
$$

which means
$$
|k x+\sqrt{3}|=2 k+\sqrt{3}-6 .
$$

From (1), we know that
$$
\begin{array}{l}
2 k+\sqrt{3}-6 \geqslant 0, \\
k \geqslant 3-\frac{\sqrt{3}}{2}
\end{array}
$$

Since $k$ is an integer, we have
$$
k \geqslant 3 \text {. }
$$

Since $k, x$ are both integers, therefore
$$
k x+\sqrt{3} \neq 0 \text {. }
$$

If $k x+\sqrt{3}>0$, then from (1) we get
$$
\begin{array}{l}
k x+\sqrt{3}=2 k+\sqrt{3}-6, \\
k(2-x)=6 .
\end{array}
$$

Noting that $k \geqslant 3$, and $k, x$ are both integers, from the above equation we get
$$
\left\{\begin{array}{l}
k=3, \\
x=0;
\end{array} \text { or } \left\{\begin{array}{l}
k=6, \\
x=1 .
\end{array}\right.\right.
$$

If $k x+\sqrt{3}<0$, then from (1) we get
$$
-(k x+\sqrt{3})=2 k+\sqrt{3}-6 \text {, }
$$

which means
$$
(2+x) k+2 \sqrt{3}-6=0 .
$$

In the above equation, only $\sqrt{3}$ is an irrational number, the rest are integers. Clearly, the above equation cannot hold.
In summary, the required integers $k$ are $k=3$ or $k=6$.