Question

Aplique as propriedades para resolver as potências:

Answer 1

1. RESOLUÇÃO
2. RESPOSTA
3. PROPRIEDADES DE POTÊNCIA

RESOLUÇÃO:

A)

${2}^{9} \times {2}^{ - 6} = {2}^{9 + ( - 6)} = {2}^{9 - 6} ={2}^{3} \: \: \: =\: \: \: 2 \times 2 \times 2\: \: \: =\: \: \: 8\\ \\ {2}^{9} \times {2}^{ - 6} ={2}^{9+ (-6)} = {2}^{3} = 8$

B)

${6}^{ - 9} \times {6}^{1} \times {6}^{7} = {6}^{( - 9) + 1 + 7} = {6}^{ - 9+ 1 + 7} = {6}^{- 1} = \frac{ {6}^{1} }{ {6}^{2} } = \frac{6 }{6 \times 6} = \frac{1}{6} \\ \\ {6}^{ - 9} \times {6}^{1} \times {6}^{7} = {6}^{( - 9) + 1 + 7} = {6}^{- 1} = \frac{1}{6}$

C)

${5}^{3} \times {5}^{ - 6} = {5}^{3 + ( - 6)} = {5}^{3 - 6} = {5}^{ - 3} = \frac{ {5}^{3} }{ {5}^{6} } = \frac{5 \times 5 \times 5}{5 \times 5 \times 5 \times 5 \times 5 \times 5} = \frac{1}{5 \times 5 \times 5 } = \frac{1}{ {5}^{3} } = \frac{1}{125} \\ \\ {5}^{3} \times {5}^{ - 6} = {5}^{3 + ( - 6)} = {5}^{ - 3} = \frac{1}{ {5}^{3} } = \frac{1}{125}$

D)

${36}^{ \frac{1}{2} } = \sqrt[2]{{36}^{1 }} = \sqrt{36} = 6$

E)

${2}^{ - 5} \times {2}^{5} \times {2}^{-1} = {2}^{( - 5) + 5 + (-1)} = {2}^{ - 5+ 5 - 1} = {2}^{- 1} = \frac{ {2}^{1} }{ {2}^{2} } = \frac{2}{2\times 2} = \frac{1}{2} = 0,5 \\ \\ {2}^{ - 5} \times {2}^{5} \times {2}^{-1} = {2}^{( - 5) + 5 + (-1)} = {2}^{- 1}=\frac{1}{2}$

F)

${7}^{ 0} = {7}^{1 - 1} = \frac{ {7}^{1} }{ {7}^{1} } = 1$

G)

${4}^{ -2} = \frac{ {4}^{1} }{ {4}^{3} } = \frac{ 4 }{ 4 \times 4 \times 4} = \frac{ 1 }{ 4 \times 4} = \frac{ 1 }{ {4}^{2} } = \frac{ 1 }{ 4 \times 4 } = \frac{ 1 }{ 16 } \\ \\ {4}^{ -2} =\frac{ 1 }{ {4}^{2} } = \frac{ 1 }{ 16 }$

H)

${2}^{ 2} \times {2}^{ - 6} = {2}^{ 2 + (-6)} = {2}^{ 2 -6} ={2}^{-4} = \frac{ 1 }{ {2}^{4} } = \frac{ 1 }{ 2 \times 2 \times 2 \times 2 } = \frac{ 1 }{ 16 }$

I)

$\frac{ {3}^{5} }{ {3}^{7} } = \frac{ 3 \times 3 \times 3 \times 3 \times 3 }{ 3 \times 3 \times 3\times 3 \times 3 \times 3 \times 3} = \frac{ 1}{ 3 \times 3 } = \frac{ 1 }{ {3}^{2} } = \frac{ 1}{ 9} \\ \\ \frac{ {3}^{5} }{ {3}^{7} } = {3}^{5-7} ={3}^{-2} = \frac{ 1 }{ {3}^{2} } = \frac{ 1}{ 9}$

J)

${ (\frac{ 2}{ 5 })}^{2} = \frac{ 2}{ 5 }\times \frac{ 2}{ 5 } = \frac{ 2 \times 2}{ 5 \times 5 } = \frac{ {2}^{2}}{ {5}^{2} } = \frac{ 4}{ 25} \\ \\ { (\frac{ 2}{ 5 })}^{2} = \frac{ {2}^{2}}{ {5}^{2} } = \frac{ 4}{ 25}$

K)

${({4}^{ 2})}^{5} = {(4 \times 4)}^{5} = (4 \times 4) \times (4 \times 4) \times (4 \times 4) \times (4 \times 4) \times (4 \times 4) = {4}^{10} \\ \\ {({4}^{ 2})}^{5} = {4}^{ 2\times 5} = {4}^{10}$

L)

${ ( \frac{ 2}{ 3 } ) }^{-3} = \frac{ \frac{ 2}{ 3 } }{ {(\frac{ 2}{ 3 })}^{4} } = \frac{ \frac{ 2}{ 3 } }{ \frac{ 2}{ 3 } \times \frac{ 2}{ 3 } \times \frac{ 2}{ 3 } \times \frac{ 2}{ 3 } } = \frac{1 }{ \frac{ 2}{ 3 } \times \frac{ 2}{ 3 } \times \frac{ 2}{ 3 } } = \frac{1}{ {(\frac{ 2}{ 3 })}^{3} } = \frac{1}{ \frac{ {2}^{3}}{ {3 }^{3} } } = \frac{1}{ \frac{ 8}{ 27 } }\: \: \:=\: \: \: \frac{1}{1 } \times \frac{ 27}{ 8 }\: \: \: =\: \: \:\frac{ 27}{ 8 } \\ \\ { ( \frac{ 2}{ 3 } ) }^{-3} = \frac{ {2}^{-3}}{ {3 }^{-3} } = \frac{ \frac{1}{{2}^{3}} }{ \frac{1}{{3}^{3}} } \: \: \: = \: \: \: \frac{1}{ {2}^{3}} \times \frac{ {3}^{3} }{1} \: \: \: = \: \: \: \frac{1}{8} \times \frac{27}{1} \: \: \:= \: \: \:\frac{27}{8} \\ \\ { (\frac{ 2}{ 3 })}^{-3} = { (\frac{ 3}{ 2})}^{3} = \frac{ {3}^{3}}{ {2}^{3} } = \frac{ 27}{ 8}$

RESPOSTA:

A)$8$

B)$\frac{1}{6}$

C)$\frac{1}{125}$

D)$6$

E)$\frac{1}{2}$

F)$1$

G)$\frac{1}{16}$

H)$\frac{1}{16}$

I)$\frac{1}{9}$

J)$\frac{4}{25}$

K)${4}^{10}$

L)$\frac{27}{8}$

PROPRIEDADES DE POTÊNCIA:

Divisão:

$\frac{{A}^{b}}{{a}^{c}} = {A}^{b-c}$

${A}^{-b} = \frac{1}{{A}^{b}}$

Multiplicação:

${A}^{b} \times {A}^{c} = {A}^{b+c}$

${({A}^{b})}^{c}={A}^{b \times c}$

${(\frac{A}{D})}^{b} =\frac{{A}^{b}}{{D}^{b}}$

Fração no Expoente:

${A}^{ \frac{b}{c} } = \sqrt[c]{{A}^{b}}$