Question

Va rog ajutati-ma la urmatoarea problema
Fie x=2,(27):5 - 2:55
y=0,6 X 0,(3) + 2 X (0,4+0,25)
z=(3,6 +5,55) X 0,12 + (3-0,12):0,036 - 754,9 X 0,02
Calculati:a) z - 2y - 33x;
b) y - x : 0,(36);
c) [0,2(6) X y - 4,4 X x]:z
  Va rog rezolvare completa.

Answer 1

/  = impartire
x=2,(27):5 - 2:55 =[(227-2)/99]/5-2/55 =(25/11)/5-2/55=5/11-2/55=
=(5×5-2)/55=23/55

y=0,6 ×0,(3) + 2 ×(0,4+0,25)=6/10 ×3/9 + 2 × (4/10+25/100)=
=3/5 ×1/3 + 2 × (2/5+1/4)=1/5 + 4/5+1/2=5/5+1/2=1+1/2=3/2

z=(3,6 +5,55) × 0,12 + (3-0,12):0,036 - 754,9 × 0,02=
=9,15 × 0,12 + 2,88:0,036 - 15,098=
=1,098 + 80 - 15,098=66


a) z - 2y - 33x =66-2×3/2-33×23/55=66-3-3×23/55=63-69/55=246/5=49,2

b) y - x : 0,(36) = 3/2 - 23/55 : 36/99 =3/2 - 23/55 × 99/36 =
=3/2 - 23/5 × 1/4 =3/2-23/20=7/20
 
c) [0,2(6) × y - 4,4 × x]:z=[(26-2)/90 ×3/2  - 44/10 × 23/55]:66=
=(24/90 ×3/2  - 22/5 × 23/55]:66=
=(12/30  - 2/5 × 23/5):66=
=(2/5  - 46/25):66=
=(10 - 46)/25:66=
=- 36/25:66=
=-6/275=-0,02(18)