8. Given that one edge of a tetrahedron is 6, and the other edges are all 5, then the radius of the circumscribed sphere of this tetrahedron is $\qquad$
$$
\begin{array}{l}
\text { Given } M C=M D=\frac{5}{2} \sqrt{3} \Rightarrow M E=M F=\frac{5}{6} \sqrt{3} \\
\Rightarrow \sin \angle C M N=\frac{3}{\frac{5}{2} \sqrt{3}}=\frac{2 \sqrt{3}}{5} \\
\Rightarrow \cos \angle C M N=\frac{\sqrt{13}}{5} \Rightarrow M O=\frac{25 \sqrt{3}}{6 \sqrt{13}},
\end{array}
$$

Therefore, the radius of the circumscribed sphere of the tetrahedron is $O B=\sqrt{\left(\frac{25 \sqrt{3}}{6 \sqrt{13}}\right)^{2}+\left(\frac{5}{2}\right)^{2}}=\frac{20 \sqrt{39}}{39}$.