б) 4x-y:(16x2-y2); 4x

б) $$\frac{4x-y}{4x} : (16x^2-y^2) = \frac{4x-y}{4x} : ((4x)^2-y^2) = \frac{4x-y}{4x} : ((4x-y)(4x+y)) = \frac{4x-y}{4x} \cdot \frac{1}{(4x-y)(4x+y)} = \frac{1}{4x(4x+y)}$$

Ответ: $$\frac{1}{4x(4x+y)}$$