• Matéria: Matemática
  • Autor: isabellysayle7
  • Perguntado 4 anos atrás

calcule as potências:

alguem me ajuda porfavor ​

Anexos:

Respostas

respondido por: thaissantos6875
1

Explicação passo-a-passo:

A)

10 {}^{ - 2}  =  \frac{1}{10 {}^{2} } \\ 10 {}^{2} = 10 \times 10 = 100 \\  \frac{1}{100}

B)

 -  (\frac{5}{8}) {}^{ - 2} = ( -  \frac{8}{5}) {}^{2}  \\ ( -  \frac{8}{5}) {}^{2} = ( \frac{8}{5}) {}^{2} =  \frac{8 {}^{2} }{5 {}^{2} } =  \frac{64}{5 {}^{2} }  \\  \frac{64}{25}

C)

(- \frac{ 3  }{ 2  }  )  ^ { -3  } = ( -  \frac{2}{3}) {}^{3} \\ ( -  \frac{2}{3}) {}^{3} =  - ( \frac{2}{3}) {}^{3} =  -  \frac{2 {}^{3} }{3 {}^{3} } =  \frac{8}{3 {}^{3} }   \\  -  \frac{8}{27}

D)

( - 3) {}^{ - 3}  =  - 3 {}^{ - 3} \\  -  \frac{1}{3 {}^{3} }  \\ 3 {}^{3} = 3 \times 3 \times 3 = 27 \\  \frac{1}{27}

E)

( +  \frac{2}{3}) {}^{ - 2} = ( \frac{3}{2}) {}^{2}  \\ ( \frac{3}{2}) {}^{2}  =  \frac{3 {}^{2} }{2 {}^{2} } =  \frac{9}{2 {}^{2} } \\  \frac{9}{4}

F)

( +  \frac{1}{2} ) {}^{ - 5}  = ( \frac{1}{2}) {}^{5}   \\ 2 {}^{5}

G)

( - 0.5) {}^{ - 3}  = 0.5 {}^{ - 3}  \\ ( \frac{1}{2}) {}^{ - 3} = 2 {}^{3} \\ 2 {}^{3}  = 2 \times 2 \times 2 = 8 \\

H)

( \frac{2}{3}) {}^{ - 1} =  \frac{3}{2}

I)

( \frac{1}{10}) {}^{2}  =  \frac{1 {}^{2} }{10 {}^{2} }  \\  \frac{1}{10 {}^{2} }  = 10 \times 10 = 100 \\  \frac{1}{100}

J)

(2.5) {}^{2}  = (  \frac{5}{2}) {}^{2}  \\  \frac{5 {}^{2} }{2 {}^{2} }  =  \frac{25}{2 {}^{2} }  \\  \frac{25}{4}

K)

( - 0.2) {}^{ - 5}  = (  -  \frac{1}{2}) {}^{ - 5}   \\ ( - 5) {}^{ 5}  =  - 5 {}^{5}  =  - (5 \times 5 \times 5 \times 5 \times 5) \\  - 3125

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