a,b,c - numere naturale
a+b+c = 6, rezulta ca a,b,c au o singura cifra
a) Astfel putem scrie forma generala a unui numar de forma [tex]\overline{a,(b)}[/tex] ca fractie ordinale, unde a si b sunt cifre in baza 10 :
[tex]\overline{a,(b)} = \frac{\overline{ab}-a}{9} = \frac{10a+b-a}{9} = \frac{9a+b}{9} = a+\frac{b}{9}[/tex]
Deci,
[tex]\overline{x,(y)} = x+\frac{y}{9}\\\\\overline{y,(z)} = y+\frac{z}{9}\\\\\overline{z,(x)} = z+\frac{x}{9}[/tex]
Daca le adunam :
[tex]\overline{x,(y)}+\overline{y,(z)}+\overline{z,(x)} = x+y+z + \frac{(x+y+z)}{9}[/tex], dar stiind ca x+y+z = 6 putem inlocui :
[tex]\overline{x,(y)}+\overline{y,(z)}+\overline{z,(x)} =6 + \frac{6}{9} = 6+\frac{2}{3} = 6+0.(6)=6.(6)[/tex]
b) In mod asemanator putem scrie forma generala a unui numar de tipul [tex]\overline{a,b(c)}[/tex] ca fractie ordinale, unde a, b, c sunt cifre in baza 10 :
[tex]\overline{a,b(c)} = \frac{\overline{abc}-\overline{ab}}{90} = \frac{100a+10b+c-10a-b}{90} = \frac{90a+9b+c}{90} = a+\frac{b}{10} +\frac{c}{90}[/tex]
Deci,
[tex]\overline{x,y(z)} = x+\frac{y}{10} + \frac{z}{90} \\\\\overline{y,z(x)} = y+\frac{z}{10} + \frac{x}{90}\\ \\\overline{z,x(y)} = z+\frac{x}{10} + \frac{y}{90}[/tex]
Daca le adunam :
[tex]suma = x+y+z+\frac{x+y+z}{10} + \frac{x+y+z}{90}[/tex], inlocuim x+y+z cu 6 si obtinem :
[tex]suma = 6+\frac{6}{10} +\frac{6}{90} = 6+0.6+0.0(6)=6.6(6)=6.(6)[/tex]
◘ Daca ai uitat transformarea fractiilor din forma zecimala in forma ordinala ai un exemplu aici :
https://brainly.ro/tema/2760639
