9.36 Given the three sides of an obtuse triangle are $3, 4, x$, then the range of $x$ is 
(A) $1<x<7$.
(B) $5<x<7$.
(C) $1<x<\sqrt{7}$.
(D) $5<x<7$ or $1<x<\sqrt{7}$.
(6th "Jinyun Cup" Junior High School Mathematics Invitational, 1989)
[Solution] Let $x$ be the longest side, and its opposite angle be $\alpha$, then $\cos \alpha25$, so $x>5$.
Also, $x4-3=1, \quad \therefore \quad 1<x<\sqrt{7}$.
Therefore, the answer is $(D)$.