Question

(CEFT-MG) Se p/q é a fraçao irredutivel equivalente a 5,666.../2,333...,o valor de p+q é igual a:

Answer 1

5, 666.......  5 inteiros 6/9  = 5 int   2/3  = ( 3 * 5) + 2 = 17/3 ****
2,333........  2 inteiros 3/9 =  2int     1/3  = ( 3 * 2) + 1 = 7/3***

p/q = 17/3 : 7/3 = 17/3 * 3/7 = 51/21 =  17/7 ***

p + q = 17/3 + 7/3 = 24/3 = 8 ****

Answer 2

.
$\frac{p}{q} \rightarrow \frac{5,666...}{2,333..} \\ \\ p= \frac{56-5}{9} = \frac{51}{9} = \frac{17}{3} \\ \\ q= \frac{23-2}{9} = \frac{21}{9} = \frac{7}{3} \\ \\ \frac{p}{q} =p\div q= \frac{17}{3} \div \frac{7}{3} = \frac{17}{3} \times \frac{3}{7} = \frac{17}{7} \\ \\ p=17 ~~~~ e~~~q=7 \\ \\ p+q=17+7=24$

Pode fazer assim também

$\frac{5,666...}{2,333...} = \frac{56-5}{23-2} = \frac{51}{21} = \frac{17}{7} \\ \\ 17+7=24$