Question

aplicando as propriedades com potências da mesma base, reduza cada expressão a uma só potência. ​Me ajudem ai é pra agora rápidoooooooo

Answer 1

Oie, Td Bom?!

a)

$= 5 {}^{4} \: . \: 5 {}^{2} \: . \: 5 {}^{6}$

$= 5 {}^{4 + 2 + 6}$

$= 5 {}^{10 + 2}$

$= 5 {}^{12}$

b)

$= ( - 7) {}^{5} \: . \: ( - 7) {}^{3} \: . \: ( - 7)$

$= ( - 7) {}^{5 + 3 + 1}$

$= ( -7) {}^{8 + 1}$

$= ( - 7) {}^{9}$

$= - 7 {}^{9}$

c)

$= 10 {}^{8} \: . \: 10$

$= 10 {}^{8 + 1}$

$= 10 {}^{9}$

d)

$= 12 {}^{20} \div 12 {}^{15}$

$= 12 {}^{20 + 15}$

$= 20 {}^{5}$

e)

$= ( - 5) {}^{9} \div ( - 5) {}^{5}$

$= ( - 5) {}^{9} \: . \: ( - 5) {}^{ - 5}$

$= ( - 5) {}^{9 - 5}$

$= ( - 5) {}^{4}$

$= 5 {}^{4}$

f)

$= 2 {}^{8} \div 2$

$= 2 {}^{8 - 1}$

$= 2 {}^{7}$

Att. Makaveli1996

Answer 2

Explicação passo-a-passo:

a)

$\sf (+5)^4\cdot(+5)^2\cdot(+5)^6=(+5)^{4+2+6}=\red{(+5)^{12}}$

b)

$\sf (-7)^5\cdot(-7)^3\cdot(-7)^1=(-7)^{5+3+1}=\red{(-7)^{9}}$

c)

$\sf (+10)^8\cdot(+10)^1=(+10)^{8+1}=\red{(+10)^{9}}$

d)

$\sf (+12)^{20}\div(+12)^{15}=(+12)^{20-15}=\red{(+12)^{5}}$

e)

$\sf (-5)^{9}\div(-5)^{5}=(-5)^{9-5}=\red{(-5)^{4}}$

f)

$\sf (+2)^{8}\div(+2)^{1}=(+2)^{8-1}=\red{(+2)^{7}}$