Question

Ejercicios con leyes de potencia

Answer 1

Respuesta:

$( - 2 {x}^{2} {y}^{3} z {)}^{ - 4} = (\frac{1}{- 2 {x}^{2} {y}^{3} z} {)}^{4} = \frac{1}{16 {x}^{8} {y}^{12} {z}^{4} }$

${x}^{9} . {x}^{4} = {x}^{9 + 4} = {x}^{13}$

${x}^{3n - 1}.{x}^{2n + 2} = {x}^{3n - 1 + 2n + 2} = {x}^{5n + 1}$

$( - 2a {b}^{2} {)}^{3} = - 8 {a}^{3} {b}^{6}$

$( - 4 {a}^{2} {b}^{3} )( - 3ab) = 12 {a}^{3} {b}^{4}$

$\frac{ {y}^{8} }{ {y}^{5} } = {y}^{8 - 5} = {y}^{3}$

$\frac{( - 3 {x}^{2} {y}^{3} z {)}^{2} }{{2 {x}^{3}y {z}^{3} }} = \frac{9 {x}^{4} {y}^{6} {z}^{2} }{{2 {x}^{3}y {z}^{3} }} = \frac{( {x}^{3} {y}^{1} {z}^{2})(x {y}^{5} )9 }{({x}^{3} {y}^{1} {z}^{2})(z)2} = \frac{9x {y}^{5}}{2z}$

${y}^{5} {y}^{8} {y}^{4} = {y}^{5 + 8 + 4} = {y}^{17}$