Division of algebraic fractions: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} : \frac{c}{d} = \frac{ad}{bc}$

Got it !

$\frac{8x-2y}{x+y} : \frac{y-4x}{4x-y} =$	$\frac{2(4x-y)}{x+y} $
$\frac{4+2a}{6-3a} : \frac{a+2}{(a-2)^2} =$	$\frac{2(2-a)}{3} $
$\frac{\frac{3c}{1-b}}{\frac{c}{3(1+b)}} =$	$\frac{9(1+b)}{1-b}$
$1-\frac{\frac{1}{x}}{x} =$	$1-1=0 $