Question 327069
((ab+b^2)/(4ab^5))/((a+b)/(6a^2b^4))
Dividing by a fraction is the same as multiplying by the fraction's reciprocal (the reciprocal is just the fraction flipped upside-down). So:
((ab+b^2)/(4ab^5))/((a+b)/(6a^2b^4))=
((ab+b^2)/(4ab^5))*((6a^2b^4)/(a+b))=
((ab+b^2)*(6a^2b^4))/((4ab^5)*(a+b))
Now you need to see if you can cancel anything in the numberator and the denominator.  If you factor the numerator (factor out a b from the first term), you can cancel out the (a+b)from the num and denom:
((ab+b^2)*(6a^2b^4))/((4ab^5)*(a+b))=
((b)(a+b)*(6a^2b^4))/((4ab^5)*(a+b))=
((b)*(6a^2b^4))/(4ab^5)=
Now multiply out the numerator to see if that will enable you to cancel anything else out:
((b)*(6a^2b^4))/(4ab^5)=
(6a^2b^5)/(4ab^5)
Now cancel the b^5:
(6a^2)/(4a)
This can also be written as (6aa)/(4a), so you can cancel an a:
(6a)/4
This can be reduced to 
3a/2
So your simplifed fraction is 3a/2.