04.36. в) (y-27)/(6-2y) + (4y)/(3-y)

в) $$ \frac{y-27}{6-2y} + \frac{4y}{3-y} = \frac{y-27}{-2(y-3)} + \frac{4y}{3-y} = - \frac{y-27}{2(y-3)} - \frac{4y}{y-3} = \frac{-(y-27) - 8y}{2(y-3)} = \frac{-y+27 - 8y}{2(y-3)} = \frac{-9y+27}{2(y-3)} = \frac{-9(y-3)}{2(y-3)} = -\frac{9}{2} $$

Ответ: $$ -\frac{9}{2} $$