Question

Ajutor ex de jos 1 b)

Answer 1

b)

[1+1,(6)-2/9+5 si 1/3]:1,1(6)=

5 si 1/3=(5×3+1)/3=16/3 ; 1,(6)=(16-1)/9=15/9 ; 1,1(6)=(116-11)/90=105/90=6/7

[1+15/9-2/9+16/3]:(6/7) =

(9+15-2+48)/9×(7/6) =

(70/9)×(7/6) = 20/3 = 20:3 = 6 intregi si 2/3

Answer 2

1, ( 6 ) = ( 16 - 1 ) / 9 = 15 / 9 = 5 / 3

5 intregi si 1 / 3 = ( 5 x 3+ 1 ) / 3 = 16 / 3

1,1( 6 ) = ( 116 - 11 ) / 90 = 105 / 90 = 21 / 18 = 7 / 6

( 1 + 5 / 3 - 2 / 9 + 16 / 3 ) : 7 / 6 =

= ( 9 / 9 + 15 / 9 - 2 / 9 + 48 / 9 ) × 6 / 7 =

= 70 / 9 × 6 / 7 =

= ( 10 × 2 ) / 3 =

= 20 / 3 = 6 intregi si 2 / 3 = 6, ( 6)