{a,b,c}~d.{0,(3);0,0(3);0,00(3)} putem sa scriem in felul urmator:
[tex] \frac{a}{0,(3)}= \frac{b}{0,0(3)}= \frac{c}{0,00(3)}=t [/tex] am egalat tot sirul cu un numar natural t.Iar acum facem in felul urmator:
[tex] \frac{a}{0,(3)}=t=\ \textgreater \ a=0,(3)\cdot t~sau~a= \frac{3}{9}t \\ \frac{b}{0,0(3)}=t=\ \textgreater \ b=0,0(3)\cdot t~sau~b= \frac{3}{90}t \\ \frac{c}{0,00(3)}=t=\ \textgreater \ c=0,00(3)\cdot t~sau~c= \frac{3}{900}t [/tex] iar acum inlocuim in relatia [tex] \frac{2a+5b}{3b+10c} [/tex] si obtinem:[tex] \frac{2\cdot \frac{3}{9}t+10\cdot \frac{3}{900}t }{3\cdot \frac{3}{90}t+10\cdot \frac{3}{900}t }= \frac{ \frac{6}{9}t+ \frac{30}{900}t }{ \frac{9}{90}t+ \frac{30}{900}t }= \frac{ \frac{630}{900}t }{ \frac{120}{900}t }= \frac{630}{120}= \frac{21}{4} [/tex]
