Question

mostre que $(\frac{3^{2} }{3^{7} }.\frac{2^{3} }{2^{2} } )^{-1} =\frac{3^{3} }{2}$

Answer 1

Resposta:

Leia abaixo

Explicação passo-a-passo:

$\boxed{(\frac{3^2}{3^7}.\frac{2^3}{2^2})^{-1} = \frac{3^3}{2}}$

$\boxed{\frac{3^7}{3^2}.\frac{2^2}{2^3} = \frac{3^3}{2}}$

$\boxed{3^5.2^{-1} = \frac{3^3}{2}}$

$\boxed{\frac{3^5}{2} \neq \frac{3^3}{2}}$