Question

o valor de x na equação x.(9/5+3/5+1/5.....)=27/4 é:

Answer 1

$(\dfrac{9}{5} + \dfrac{3}{5}+\dfrac{1}{5}+...) = S \ onde \ S \ a \ soma \ dos \ infinitos \ termos \ uma \ PG \\ \\ \ cuja \ \text{raz{\~ a}o} \ \textit{\' e} \ q, \ |q| \ \textless \ 1 \ e \ a_1 \neq 0. \\ \\ PG: (\dfrac{9}{5} , \dfrac{3}{5},\dfrac{1}{5},...) \\ \\ \text{Raz{\~ a}o} = q = \dfrac{ \dfrac{3}{5} }{ \dfrac{9}{5}}= \dfrac{3}{5} *\dfrac{5}{9} = \dfrac{1}{3}$

$S= \dfrac{ \dfrac{9}{5} }{1- \dfrac{1}{3} } = \dfrac{ \dfrac{9}{5} }{ \dfrac{2}{3}} = \dfrac{9}{5} * \dfrac{3}{2} = \dfrac{27}{10} \\ \\ x * \dfrac{27}{10} = \dfrac{27}{4} \Longleftrightarrow \dfrac{x}{10} = \dfrac{1}{4} \Longleftrightarrow 4x = 10 \Longleftrightarrow x = \dfrac{10}{4} = \dfrac{5}{2} \\ \\ \therefore x = \dfrac{5}{2}$ 

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Obrigado pela oportunidade. 
Boa sorte, bons estudos.
SSRC - 2015 
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