Question

Hi. Ayuda xfavor :c
con procedimiento

Answer 1

Hola! :3
Recordemos que:
0,aaa... = a/9
0,b = b/10
x y/z = [x(z) + y]/z
Ejemplo:
0,111 = 1/9
0,2 = 2/10 = 1/5
1 1/2 = [1(2) + 1]/2 = (2 + 1)/2 = 3/2
Con esto, resolvemos:
14) (0,888... ÷ 0,555... - 0,666... ÷ 0,777...) × 2 9/13 + 3 1/2 ÷ 0,5
(8/9 ÷ 5/9 - 6/9 ÷ 7/9) × [3(2) + 1]/2 ÷ 5/10
(8/9 × 9/5 - 6/9 × 9/7) × (6 + 1)/2 ÷ 1/2
[(8 × 9)/(9 × 5) - (6 × 9)/(9 × 7)] × 7/2 ÷ 1/2
(72/45 - 54/63) × 7/2 × 2/1
Simplificando:
(8/5 - 6/7) × 7/1 × 1/1
[8(7) - 5(6)]/[5(7)] × 7
(56 - 30)/35 × 7
Otra vez simplificamos:
(56 - 30)/5 × 1
26/5
25/5 + 1/5
5 + 1/5
5 1/5

15) (3,222... - 2,888...) × 3 + (0,777... - 0,555...) ÷ 2 1/2
(3 + 0,222... - 2 - 0,888...) × 3 + (7/9 - 5/9) ÷ [2(2) + 1]/2
(1 + 2/9 - 8/9) × 3 + (2/9) ÷ (4 + 1)/2
(1 - 6/9) × 3 + (2/9) ÷ 5/2
(9/9 - 6/9) × 3 + 2/9 × 2/5
(3/9) × 3 + (2 × 2)/(9 × 5)
(3 × 3)/9 + 4/45
9/9 + 4/45
1 + 4/45
1 4/45

Espero haberte ayudado :D
Atte: Jean07122006