Question

Sendo x= 4,333... e y= 3,444..., calcule o valor de:

A) x.y/×+y

B) x:y/x+y

Answer 1

1° Passo: Devemos transformar as dízimas periódicas em frações, para isso:

x= 4,333... (I)

10.x= 43,333... _(II)

(I) - (II)

10.x - x = 43,333... - 4,333...

9x = 39

x= 39/9

y= 3,444... (III)

10.y=34,444... (IV)

(III) - (IV)

10.y-y = 34,444...-3,444...

9y= 31

y= 31/9

Achamos os valores de x e y, agora vamos fazer o cálculo das alternativas solicitadas

a)

$\frac{x.y}{x+y} = \frac{\frac{39}{9}.\frac{31}{9}}{\frac{39}{9}+\frac{31}{9}} = \frac{\frac{1209}{81} }{\frac{70}{9}} = \frac{1209.9}{70.81} \div 9 = \frac{1209}{70.9} \div 3 = \frac{403}{70.3} = \boxed{\frac{403}{210}}$

b)

$\frac{x\div y}{x+y} = \frac{\frac{39}{9}\div \frac{31}{9}}{\frac{39}{9}+\frac{31}{9}} = \frac{\frac{39}{9}.\frac{9}{31}}{\frac{70}{9}} = \frac{\frac{39}{31} }{\frac{70}{9} }= \frac{39.9}{31.70} =\boxed{\frac{351}{2170}}$

Answer 2

Bom dia!

x = 4,3333... = 4 + 0,3333... = 4 + 3/9 = 4 + 1/3 = 12/3 + 1/3 = 13/3

y = 3,4444... = 3 + 0,444... = 3 + 4/9 = 27/9 + 4/9 = 31/9

a) (13/3 . 31/9) / (13/3 + 31/9) = (403/27)/(39/9 + 31/9) = (403/27)/(70/9) = 403/27 . 9/70 = 3627/1890 = 1209/603 = 403/201

b) (13/3)/(31/9) / (13/3 + 31/9) = (13/9 . 9/31)/(39/9 + 31/9) = (117/279)/(70/9) = 117/279 . 9/70 = 1053/19530 = 351/6510

Espero ter ajudado!