8. Let one edge of a tetrahedron be 6, and the other edges all be 5. Then the radius of the circumscribed sphere of this tetrahedron is $\qquad$
8. $\frac{20 \sqrt{39}}{39}$.

In the tetrahedron $P A B C$, let
$$
B C=6, P A=P B=P C=A B=A C=5 \text {. }
$$

Let the midpoint of $B C$ be $D$, and the circumcenter of $\triangle A B C$ be $E$.
Then $B D=C D=3$,
$$
A E=B E=C E=\frac{25}{8}, D E=\frac{7}{8} \text {. }
$$

Since $P A=P B=P C=5$, we have
$P E \perp$ plane $A B C$,
$$
P E=\sqrt{P A^{2}-A E^{2}}=\frac{5 \sqrt{39}}{8} \text {. }
$$

Clearly, the center of the circumsphere of the tetrahedron $O$ lies on the segment $P E$, and let the radius of the circumsphere be $r$. Then
$$
O P=O A=r, O E=P E-r \text {. }
$$

In the right triangle $\triangle O A E$, by the Pythagorean theorem we get
$$
\begin{array}{l}
O A^{2}=O E^{2}+A E^{2} \\
\Rightarrow r^{2}=(P E-r)^{2}+A E^{2} \\
\Rightarrow r=\frac{P E^{2}+A E^{2}}{2 P E}=\frac{20 \sqrt{39}}{39} .
\end{array}
$$