Question

A soma das raízes da equação abaixo é:

preciso de urgência ​

Answer 1

Resposta:

7^(1+x) +1/7^(x) =8

7 *7^(x) +1/7^(x)= 8

tudo vezes 7^(x)

7 * 7^(x) * 7^(x) + 1 =8*7^(x)

7 * 7^(2x)  -8*7^(x) + 1=0

faça y=7^(x)

7y² -8y+1=0

y'=[8+√(64-28)]/14 =(8 +6)/14 = 1

y''=[8-√(64-28)]/14 =(8 -6)/14 = 1/7  

y=1=7^(x) ==>x=0

y=1/7=7^(-1) =7^(x)  ==>x=-1

soma = 0 -1 = -1

Answer 2

$7 {}^{1 + x} + \frac{1}{7 {}^{x} } = 8$

$7 {}^{1} \: . \: 7 {}^{x} + \frac{1}{7 {}^{x} } = 8$

$7 \: . \: 7 {}^{x} + \frac{1}{7 {}^{x} } = 8$

$\boxed{7 {}^{x} ⟷t}$

$7t + \frac{1}{t} = 8$

$7t + \frac{1}{t} - 8 = 0$

$\frac{7t {}^{2} + 1 - 8t }{t} = 0$

$7t {}^{2} + 1 - 8t = 0$

$7t {}^{2} - 8t + 1 = 0$

$7t {}^{2} - t - 7t + 1 = 0$

$t \: . \: (7t - 1) - (7t - 1) = 0$

$(7t - 1) \: . \: (t - 1) = 0$

$7t - 1 = 0⟶7t = 1⟶\boxed{t = \frac{1}{7} }$

$t - 1 = 0⟶\boxed{t = 1 }$

$\boxed{t⟷7 {}^{x} }$

$7 {}^{x} = \frac{1}{7} ⟶7 {}^{x} = 7 {}^{ - 1} ⟶\boxed{x = - 1}$

$7 {}^{x} = 1 ⟶7 {}^{x} = 7 {}^{0} ⟶\boxed{x = 0}$

$S= - 1 + 0 = \boxed{-1}$

Att. YRZ