Question

Al simplificar $\frac{4 ^{-2}+ 4^{-1} }{ 4^{2} }$ se obtiene?

Answer 1

${ \frac{4^{-2}+4^{-1}}{4^2}}\\\\{ \frac{( \frac{1}{4})^2+ \frac{1}{4} }{16}}\\\\{ \frac{ \frac{1}{16}+ \frac{1}{4} }{16}}\\\\{ \frac{ \frac{5}{16} }{16}}\\\\{ \frac{5}{256}}\\\\{salu2. :)}$