18. Solve the equation $\sqrt{x^{2}-4}+2 \sqrt{x^{2}-1}=\sqrt{7} x$.
18. In a circle with diameter $A B=x>2$, take points $C$ and $D$ on the two semicircles such that $A C=2, A D=1$, then $B C=\sqrt{x^{2}-4}, B D=\sqrt{x^{2}-1}$. By Ptolemy's theorem, we have $\sqrt{x^{2}-4}+2 \sqrt{x^{2}-1}=C D \cdot x$. Comparing with the original equation, we get $C D=\sqrt{7}$. In $\triangle A C D$, by the cosine rule, we have $\cos \angle C A D=-\frac{1}{2}$, so $\angle C A D=120^{\circ}$. Therefore, $x=\frac{C D}{\sin \angle C A D}=\frac{2}{3} \sqrt{21}$.