Question

ayudenme estoy en un examen.

Answer 1

Respuesta:

$\frac{\left(3^2\right)^3\cdot \left(3^4\right)^5\cdot \:3^3}{\left(3^3\right)^6\cdot \:3^4}$

$\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \frac{x^a}{x^b}=\frac{1}{x^{b-a}}\\\\=\frac{\left(3^4\right)^5\left(3^2\right)^3}{3^{4-3}\left(3^3\right)^6}\\\\=\frac{\left(3^4\right)^5\left(3^2\right)^3}{3\left(3^3\right)^6}$

$=\frac{\left(3^4\right)^5\cdot \left(3^2\right)^3}{3\cdot \left(3^3\right)^6}$

$\left(3^4\right)^5==3^{4\cdot \:5}=3^2^0$

$=\frac{3^{20}\left(3^2\right)^3}{3\left(3^3\right)^6}\\\\=\frac{3^{19}\left(3^2\right)^3}{\left(3^3\right)^6}\\\\=\frac{3^{19}\left(3^2\right)^3}{3^{18}}\\\\=\frac{3^{19}\cdot \:3^6}{3^{18}}$

$3^{19}\cdot \:3^6\\\\\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^b\cdot \:a^c=a^{b+c}\\\\=3^{19+6}\\\\=3^{25}\\\\=\frac{3^{25}}{3^{18}}\\\\\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\=3^{25-18}\\\\=3^7\\\\=2187$