16. Given $\sin \left(\frac{\pi}{4}-x\right)=\frac{5}{13}$, and $x \in\left(0, \frac{\pi}{4}\right)$, find the value of $\frac{\cos 2 x}{\cos \left(\frac{\pi}{4}+x\right)}$.
16. Let $\frac{\pi}{4}-x=\alpha$, then $\alpha \in\left(0, \frac{\pi}{4}\right), \sin \alpha=\frac{5}{13}, \cos \alpha=\frac{12}{13}$.
$$
\begin{aligned}
\cos 2 x & =\cos \left[2\left(\frac{\pi}{4}-\alpha\right)\right]=\cos \left[\frac{\pi}{2}-2 \alpha\right]=\sin 2 \alpha=2 \sin \alpha \cos \alpha, \cos \left(\frac{\pi}{4}+x\right)= \\
\cos \left(\frac{\pi}{2}-\alpha\right) & =\sin \alpha, \text{ so } \frac{\cos 2 x}{\cos \left(\frac{\pi}{4}+x\right)}=2 \cos \alpha=\frac{24}{13} .
\end{aligned}
$$