Question

Calculați:
a)3/11+0,(1)+0,2(32)
b)1,(18)+0,41(6)+0,25
c)1,1(2)+2,2(3)+3,3(4)
Dau 54 puncte

Answer 1

a)
3/11+0,(1)+0,2(32) =
= 3/11 + 1/9 + (232 - 2)/990 =
= 3/11 + 1/9 + 230/990 =
= 3/11 + 1/9 + 23/99 =
= 3×9/99 + 11/99 + 23/99 =
= (27 + 11 + 23)/99 =
= 61/99 =
= 0,(61)

b)
1,(18)+0,41(6)+0,25 =
= (118 -1)/99 + (416-41)/900 + 25/100 =
= 117/99+ 375/900 + 1/4 =
= 13/11 + 15/36 + 1/4 =
= 13×36/396 + 11×15/396 + 99/396 =
= 468/396 + 165/396 + 99/396 =
= 732/396 =
= 1,(84)

c)
1,1(2)+2,2(3)+3,3(4) =
= (112 - 11)/90 + (223-22)/90 + (334-33)/90 =
= 101/90 + 201/90 + 301/90 =
= (101 + 201 + 301)/90 =
= 603/90 =
= 6,7