Question

questão 19 s=(raiz quadrada de x)

Answer 1

lembrando que:

$\boxed{a^{ \frac{x}{y} } = \sqrt[y]{a^x}}$

$\boxed{\boxed{a^1 * a^2 = a^{1+2}}}$

aplicando isso temos
$A = x^{ \frac{1}{3} } *(x ^{ \frac{1}{3} })^{ \frac{1}{3} } * ((x^{ \frac{1}{3} })^{ \frac{1}{3} })^{ \frac{1}{3} }...\\\\\boxed{\boxed{A= x^{ \frac{1}{3} } *x ^{ \frac{1}{3^2} }* x^{ \frac{1}{3^3} }....}}$

observe que temos
$\boxed{\boxed{A = x^{( \frac{1}{3}+ \frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^n} )}}}$

1/3 + 1/3² + 1/3³ .... é a soma de uma PG infinita

a1 = 1/3
q = 1/3

$S_n = \frac{a_1}{1-q} = \frac{ \frac{1}{3} }{1- \frac{1}{3} }= \frac{ \frac{1}{3} }{ \frac{3-1}{3} }= \frac{1}{2}$

portanto teremos
$A = x^{( \frac{1}{3}+ \frac{1}{3^2}+\frac{1}{3^3}+..... )}}}=x^{ \frac{1}{2} }= \sqrt[]{x}$