Question

alguém pode me ajudar?​

Answer 1

1)

a) (3^4)*(3^5) = 3^(4 + 5) = 3^9

b) ((x^3)^4)/9 = (x^(3*4))/9 = (x^12)/9

c) (7^9)/(7^4) = 7^(9 - 4) = 7^5

d) (10^12)/(10^5) = 10^(12 - 5) = 10^7

e) a^(n + 1)*a^(n - 2) = a^(n + 1 + (n - 2)) = a^(n + 1 + n - 2) = a^(2n - 1)

f) (10^3)^2 = 10^(3*2) = 10^6

2)

$\frac{3^{12} - 3^{11} - 3^{10}}{3^{11} + 3^{10} + 3^{10}} =\\\\\frac{3^{10}(3^2 - 3^1 - 3^0)}{3^{10}(3^1 + 3^0 + 3^0)} =\\\\\frac{3^2 - 3^1 - 1}{3^1 + 3^0 + 3^0} =\\\\\frac{9 - 3 - 1}{3 + 1 + 1} = \\\\\frac{5}{5} = \\\\1$