Question

Hola, necesito ayuda en este ejercicio para mi, muy complicado, si no sabes no respondas plss ❤️❤️:(​

Answer 1

Respuesta:   $2^{25x^2-121}$

Explicación paso a paso:

$\left(\frac{64^{2x-3}}{128^{x-1}}\right)^{5x+11}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}$

$\left(\frac{64^{2x-3}}{128^{x-1}}\right)^{5x+11}=\frac{\left(64^{2x-3}\right)^{5x+11}}{\left(128^{x-1}\right)^{5x+11}}$

$=\frac{\left(64^{2x-3}\right)^{5x+11}}{\left(128^{x-1}\right)^{5x+11}}$

$\mathrm{Simplificar}\:\left(64^{2x-3}\right)^{5x+11}:\quad 64^{\left(2x-3\right)\left(5x+11\right)}$

$=\frac{64^{\left(2x-3\right)\left(5x+11\right)}}{128^{\left(x-1\right)\left(5x+11\right)}}$

$\mathrm{Factorizar}\:64^{\left(2x-3\right)\left(5x+11\right)}:\quad 2^{6\left(2x-3\right)\left(5x+11\right)}$

$\mathrm{Factorizar}\:128^{\left(x-1\right)\left(5x+11\right)}:\quad 2^{7\left(x-1\right)\left(5x+11\right)}$

$=\frac{2^{6\left(2x-3\right)\left(5x+11\right)}}{2^{7\left(x-1\right)\left(5x+11\right)}}$

$\mathrm{Cancelar\:}\frac{2^{6\left(2x-3\right)\left(5x+11\right)}}{2^{7\left(x-1\right)\left(5x+11\right)}}:\quad 2^{25x^2-121}$

$=2^{25x^2-121}$