Folklore

Solve the equation: $2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}=\frac{10}{x-4}$.
Since the left, and therefore the right, part of the equation takes only positive values, then $x>4$. On the interval $(4,+\infty)$, the function

$f(x)=2 \sqrt{x^{2}-16}+\sqrt{x^{2}-9}$ is increasing, while the function $g(x)=\frac{10}{x-4}$ is decreasing, so the equation $f(x)=g(x)$ has no more than one root.

It remains to note that $f(5)=g(5)$.

## Answer

$x=5$