Question

AAAAAAYYYYUUDDDDAAAAAA PORFASSSSSSSSS SOLO EXPERTOS $M=(9(\sqrt[K]{3} )^{K-12} . (\sqrt[K]{27} )^{K+4} )^{-\frac{1}{3} }$

Answer 1

Respuesta:

1/9

Explicación paso a paso:

$(3^{2} *3^{\frac{1}{K}*K-12 } *27^{\frac{1}{K}*K+4 })^{-\frac{1}{3} }$

$(3^{2} *3^{-11 } *27^{5 })^{-\frac{1}{3} }$

$(3^{-9}*3^{15} )^{-1/3}$

$(3^{6})^{-\frac{1}{3} }$

$\sqrt[3]{\frac{1}{3^{6} } }$

simplificando

$\frac{1}{3^{2} }$

$\frac{1}{9}$