• Matéria: Matemática
  • Autor: 00001101304613sp
  • Perguntado 6 anos atrás

Exercícios de potência.​

Anexos:

Respostas

respondido por: Anônimo
1

Explicação passo-a-passo:

a)

\sf 2^3\cdot2^4\cdot2^6\cdot2^2=2^{3+4+6+2}=2^{15}

b)

\sf (6^3)^6=6^{3\cdot6}=6^{18}

c)

\sf \dfrac{6^7}{6^5}=6^{7-5}=6^2

d)

\sf (3^3)^4=3^{3\cdot4}=3^{12}

e)

\sf \dfrac{11^5}{11^3}=11^{5-3}=11^2

f)

\sf \left[\left(\dfrac{1}{2}\right)^2\right]^{-1}=\left(\dfrac{1}{2}\right)^{2\cdot(-1)}=\left(\dfrac{1}{2}\right)^{-2}

g)

\sf \left(\dfrac{1}{4}\div\dfrac{1}{2}\right)^{-3}

\sf =\left(\dfrac{2^{-2}}{2^{-1}}\right)^{-3}

\sf =[2^{-2-(-1)}]^{-3}

\sf =(2^{-2+1})^{-3}

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

\sf =2^{(-1)\cdot(-3)}

\sf =2^{3}

h)

\sf (3\cdot4)^2=12^2

i)

\sf \left(\dfrac{2}{5}\right)^{-1}\cdot\left(\dfrac{2}{5}\right)^{3}=\left(\dfrac{2}{5}\right)^{-1+3}=\left(\dfrac{2}{5}\right)^{2}

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