Question

$\frac{y^{2}-16 }{4y} *\frac{2y+8}{y^{2}+8a+16 }$

Answer 1

Ответ: $\frac{(y-4)(y+4)}{4y}*\frac{2y+8}{(y+4)(y+4)}$ - сокращаем: $\frac{(y-4)}{4y}*\frac{2y+8}{(y+4)} = \frac{2y^{2}-8y+8y-32}{4y^{2}+16 }$ =

$\frac{2(y^{2} -16)}{2(2y^{2}+8 )} = \frac{y^{2} -16}{2y^{2}+8}$

Объяснение: