Question

Calcule as potencias

Answer 1

a.

$\boxed{{4}^{ - 3} = \frac{1}{ {4}^{3} } = \red{ \frac{1}{64} }}$

b.

$\boxed{ {6}^{ - 2} = \frac{1}{ {6}^{2} } = \red{ \frac{1}{36}} }$

c.

$\boxed{ {1}^{ - 30} = \frac{1}{ {1}^{30} } =\red{ \frac{1}{1} = 1}}$

d.

$\boxed{{10}^{ - 4} = \frac{1}{ {10}^{4} } =\red{ \frac{1}{10 \: 000}}}$

e.

$\boxed{ {0.2}^{ - 3} = \frac{1}{0.2 ^{3} } = \red{ \frac{1}{0.008}}}$

Answer 2

$\Large\boxed{\begin{array}{l}\underline{\rm Pot\hat encia~de~expoente~negativo}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\bigg(\dfrac{b}{a}\bigg)^{-n}=\bigg(\dfrac{a}{b}\bigg)^n}}}}\end{array}}$

$\boxed{\begin{array}{l}\tt a)~\sf4^{-3}=\bigg(\dfrac{4}{1}\bigg)^{-3}=\bigg(\dfrac{1}{4}\bigg)^3=\dfrac{1}{64}\\\\\tt b)~\sf 6^{-2}=\bigg(\dfrac{6}{1}\bigg)^{-2}=\bigg(\dfrac{1}{6}\bigg)^2=\dfrac{1}{36}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)\\\sf 1^{a}=1~\forall~a\in\mathbb{R}\\\sf portanto~1^{-30}=1\end{array}}$

$\large\boxed{\begin{array}{l}\tt d)~\sf 10^{-4}=\bigg(\dfrac{10}{1}\bigg)^{-4}=\bigg(\dfrac{1}{10}\bigg)^4=\dfrac{1}{10~000}\end{array}}$

$\large\boxed{\begin{array}{l}\tt e)~\sf0,2=\dfrac{2}{10}=\dfrac{1}{5}\\\\\sf0,2^{-3}=\bigg(\dfrac{1}{5}\bigg)^{-3}=\bigg(\dfrac{5}{1}\bigg)^3=5^3=125\end{array}}$