Respuesta :

choixongdong
   4y(1 - 2y)
= --------------
   5(2y - 1)

   -4y(2y - 1)
= --------------
    5(2y - 1)

= -4y / 5

hope it helps
Banabanana
[tex] \cfrac{4y-8y^2}{10y-5}= \cfrac{-(8y^2-4y)}{5(2y-1)}= \cfrac{-4y(2y-1)}{5(2y-1)}= -\cfrac{4y}{5} [/tex]