若 $\sin x \cdot \cos x=\frac{1}{8}$ 且 $\frac{\pi}{4}<x<\frac{\pi}{2}$ ，则 $\cos x-\sin x$ 的值是( )
$\because \frac{\pi}{4}<x<\frac{\pi}{2}, \therefore \cos x-\sin x<0 ，$

$\therefore(\cos x-\sin x)^{2}=1-2 \sin x \cos x=1-2 \times \frac{1}{8}=\frac{3}{4}$,

$\therefore \cos x-\sin x=-\frac{\sqrt{3}}{2}$. 故选 C.