Question

áaaaaaayuda es un gran deber para hoy ​

Answer 1

Explicación paso a paso:

para resolver éstas divisiones utilizaremos factor común:

a)

$\frac{ {x}^{7} }{ {x}^{5} } = \frac{( {x}^{2})( {x}^{5} ) }{x {}^{5} } = \frac{x {}^{2}( \not \: x {}^{5}) }{ \not \: x {}^{5} } = {x}^{2}$

b)

$\frac{6x {}^{3}y {}^{2} }{2y} = \frac{2y(3x {}^{3}y) }{2y} = \frac{ \not{2y(3x {}^{3}y) }}{ \not2y} = 3x {}^{3} y$

c)

$\frac{21x {}^{2}y {}^{3} }{7x {}^{2}y {} {}^{2}} = \frac{(7 {x}^{2}y {}^{2})3y }{7 {x}^{2} {y}^{2} } = \frac{ \not(7 { \not \: x}^{2} \not \: y {}^{2}) 3y }{ \not7 { \not \: x}^{2} { \not \: y}^{2} } = 3y$

d

$\frac{9a {}^{2} - 6a }{3a} = \frac{3a(3a - 2)}{3a} = \frac{ \not3a(3a - 2)}{ \not3a} = 3a - 2$

e)

$\frac{10 {a}^{3} + 8}{2} = \frac{2(5 {a}^{3} + 4)}{2} = \frac{ \not2(5 {a}^{3} + 4) }{ \not2} = 5 {a}^{3} + 4$

f)

$\frac{12 {a}^{2} + 8a + 24 }{2} = \frac{2(6a {}^{2} + 4a + 12) }{2} \\ = \frac{ \not2(6a {}^{2} + 4a + 12) }{ \not2} = 6 {a}^{2} + 4a + 12$

espero que sea de tu ayuda

Saludos :)