Question

Cum se calculeaza  calcullele cu perioada
0,(5)+0,(4)=
0,(35)+0,(13)=
0,(7)+0,(51)=
0,(75)+2,(5)=
12,3(4)+0,5(6)=
2,5(15)+3,(21)=
0,(135)+0,531=
7,35+1 pe 8 +2,(3)=
2 intregi si 1 pe 3 +0,25+0,(5)=
0,45+2,(3)+3,(15)+5,55=

Answer 1

0,(5)+0,(4) = 5/9+4/9 = 9/9 =1

0,(35)+0,(13) = 35/99+13/99 = 48/99 = 16/33 

0,(7)+0,(51) = 7/9+51/99 = 77/99+51/99 = 128/99 

0,(75)+2,(5) = 75/99+(25-2)/9 = 75/99+23/9 =75/99+253/99 = 328/99

12,3(4)+0,5(6) = (1234-123)/90+(56-5)/90 = 1111/90+51/90 = 1162/90 = 581/45 

2,5(15)+3,(21) = (2515-25)/990+(321-3)/99 = 2490/990+318/99 = 249/99+318/99 = 567/99 = 189/33 = 63/11

0,(135)+0,531 = 135/999+531/1000 = 135000/999000+530469/999000 = 665469/999000 = 73941/111000 = 24647/37000 

7,35+1/8+2,(3) = 735/100+1/8+(23-2)/9 = 735/100+1/8+21/9 = 147/20+1/8+7/3 = 1764/240+30/240+560/240 = 2354/240 = 1177/120

2 intregi 1/3+0,25+0,(5) = 7/3+25/100+5/9 = 7/3+5/20+5/9 = 420/180+45/180+100/180 = 565/180 = 113/36

0,45+2,(3)+3,(15)+5,55 = 45/100+(23-2)/9+(315-3)/99+555/100 = 45/100+21/9+312/99+555/100 = 9/20+7/3+104/33+111/20 = 297/660+1540/660+2080/660+3663/660 = 7580/660 = 758/66 = 379/33