Question

Ayudenme con esto por favor :)

Answer 1

Respuesta:

1.

$\frac{ {b}^{2} + 4bc}{ {b}^{3} + 64 {c}^{3} } \\ \frac{b(b + 4c)}{(b + 4c)( {b}^{2} - 4bc + 16 {c}^{2} )} \\ = \frac{b}{ {b}^{2} - 4bc + 16 {c}^{2} }$

2.

$\frac{ {b}^{2} + 4bc}{ {b}^{2} - 16 {c}^{2} } \\ \frac{b(b + 4c)}{(b - 4c)(b + 4c)} \\ = \frac{b}{b - 4c}$

3.

$\frac{ {b}^{2} - 4bc}{ {b}^{3} - 64 {c}^{3} } \\ \frac{b(b - 4c)}{(b - 4c)( {b}^{2} + 4bc + 16 {c}^{2} )} \\ = \frac{b}{ {b}^{2} + 4bc + 16 {c}^{2} }$

4.

$\frac{ {b}^{2} - 16 {c}^{2} }{ {b}^{2} - 8bc + 16 {c}^{2} } \\ \frac{(b - 4c)(b + 4c)}{(b - 4c)(b - 4c)} \\ = \frac{b + 4c}{b - 4c}$

5.

$\frac{ {b}^{2} - 16 {c}^{2} }{ {b}^{2} + 8bc - 16 {c}^{2} } \\ \frac{(b - 4c)(b + 4c)}{ {(b + 4c)}^{2} } \\ = \frac{b - 4c}{b + 4c}$

Espero haberte ayudado :)

Éxitos.

Answer 2

Espero haberte ayudado.

:))