Question

sejam x e y dois números reais, sendo x= 2, 333... e y = 0,1212..., dizimas periodicas. a soma das frações geratrizes de x e y é:​

Answer 1

Resposta:

.      27/11

Explicação passo-a-passo:

.

.     x  =  2,333...    e     y  =  0,121212...

.

.     x  +  y  =  2,333...  +  0,121212...

.                 =  2  +  3/9  +  12/99

.                 =  2  +  1/3   +  4/33

.                 =  7/3  +  4/33                    (m.m.c.  =  33)

.                 =  77/33  +  4/33

.                 =  (77  +  4)/33

.                 =  81/33                     (simplifica por 3)

.                 =  27/11

.

(Espero ter colaborado)

Answer 2

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$\large\boxed{\begin{array}{l}\sf x=2,333...\cdot10\\\sf10x=23,333...\\-\underline{\begin{cases}\sf 10x=23,\diagup\!\!\!3~\diagup\!\!\!3~\diagup\!\!\!3....\\\sf x=2,\diagup\!\!\!3~\diagup\!\!\!3~\diagup\!\!\!3...\end{cases}}\\\sf 9x=21\\\sf x=\dfrac{21\div3}{9\div3}\\\sf x=\dfrac{7}{3}\end{array}}$

$\boxed{\begin{array}{l}\sf y=0,1212...\cdot100\\\sf 100y=12,1212...\\-\underline{\begin{cases}\sf 100y=12,\diagdown\!\!\!\!\!\!12~\diagdown\!\!\!\!\!\!12~\diagdown\!\!\!\!\!\!12...\\\sf y=0,\diagdown\!\!\!\!\!\!12~\diagdown\!\!\!\!\!\!12~\diagdown\!\!\!\!\!\!12...\end{cases}}\\\sf 99y=12\\\sf y=\dfrac{12\div3}{99\div3}\\\sf y=\dfrac{4}{33}\end{array}}$

$\large\boxed{\begin{array}{l}\sf x+y=\dfrac{7}{3}+\dfrac{4}{33}\\\sf x+y=\dfrac{77+4}{33}\\\sf x+y=\dfrac{81\div3}{33\div3}\\\sf x+y=\dfrac{27}{11}\end{array}}\blue{\checkmark}$