Not necessarily. Assume that $ad-bc < 0$ and $cx > d$. Then
$$x < 1 \implies (ad-bc)x > (ad-bc) \implies -ad - bcx > -bc - adx$$
$$\implies acx - ad - bcx + bd > acx -bc-adx + bd \implies (cx-d)(a-b) > (ax-b)(c-d)$$
$$\frac{a-b}{c-d} > \frac{ax-b}{cx-d}$$