7. In tetrahedron $ABCD$, one edge has a length of 3, and the other five edges have a length of 2. Then the radius of its circumscribed sphere is $\qquad$ .
Answer: $\frac{\sqrt{21}}{3}$.
Solution: Let $B C=3, A B=A C=A D=B D=C D=2$, $E, F$ are the midpoints of $B C, A D$ respectively, and $D$'s projection on plane $A B C$ is $H$.

$H$ should be the circumcenter of $\triangle A B C$, since any point on $D H$ is equidistant from $A, B, C$, the center of the circumscribed sphere $O$ is on $D H$. Because $\triangle A B C \cong \triangle D B C$, we have $A E=D E$, so $E F$ is the perpendicular bisector of $A D$, thus the center of the sphere $O$ is the intersection of $D H$ and $E F$, and is the orthocenter of $\triangle E A D$. Let the radius be $r$, from $\triangle D O F \sim \triangle E A F$, we get $r=\frac{A E \cdot D F}{E F}$. And $A E^{2}=2^{2}-\left(\frac{3}{2}\right)^{2}=\frac{7}{4}, D F=A F=1, E F^{2}=A E^{2}-A F^{2}=\frac{3}{4}$, so $r=\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3}$.