Question

1.Transforma in fractii ordinare ireductibile fractiile periodice : 0,(3) ; 1,1(3) ; 0,01(05)
2.Reolva ecuatiile:
a) 5,2+x=14,6-x
b) 6x+3,5=21,5

Answer 1

0,(3)=3/9=1/3
1,1(3)=(113-11)/90=102/90=34/30=17/15
0,01(05)=(105-1)/9900=104/9900=26/2475

2.
a) 5,2+x=14,6-x
2x=14,6-5,2
2x=9,4
x=4,7

b) 6x+3,5=21,5
6x=21,5-3,5
6x=18
x=3

Answer 2

0,(3)= 3/9= 1/31,1(3)=(113-11)/90=17/15
0,01(05)=(105-1)/9900=26/2475

5,2+x=14,6-x
2x=14,6-5,2
x=4,7

6x+3,5=21,5
6x=21,5-3,5
x=3