Question

Помогите пожалуйста решить пример, 8 класс.

Answer 1

$\frac{y}{y-4} +\frac{y^2-2y+12}{16-y^2} =\frac{y}{y-4} -\frac{y^2-2y+12}{y^2-16} =\frac{y(y+4)}{(y-4)(y+4)} -\frac{y^2-2y+12}{(y-4)(y+4)}\\ =\frac{y^2+4y-y^2+2y-12}{(y-4)(y+4)} =\frac{6y-12}{(y-4)(y+4)}= \frac{6(y-2)}{y^2-16}$

Answer 2

y/y-4 + y^2-2y+12/16-y^2

y/y-4 + y^2-2y+12/(4-y)(4+y)

y/y-4 - y^2-2y+12/(y-4)(4+y)

y(4+y)-(y^2-2y+12)/(y-4)(4+y)

4y+y^2-y^2+2y-12/y^2-16

6y-12/y^2-16