Question

Știind că x+y+z=6,calculeaza : a)x,(y)+y,(z)+z,(x) b)x,y(z)+y,z(x)+z,x(y)

Answer 1

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 $\overline{a,(b)}$ ca fractie ordinale, unde a si b sunt cifre in baza 10 :

$\overline{a,(b)} = \frac{\overline{ab}-a}{9} = \frac{10a+b-a}{9} = \frac{9a+b}{9} = a+\frac{b}{9}$

Deci,

$\overline{x,(y)} = x+\frac{y}{9}\\\\\overline{y,(z)} = y+\frac{z}{9}\\\\\overline{z,(x)} = z+\frac{x}{9}$

Daca le adunam :

$\overline{x,(y)}+\overline{y,(z)}+\overline{z,(x)} = x+y+z + \frac{(x+y+z)}{9}$, dar stiind ca x+y+z = 6 putem inlocui :

$\overline{x,(y)}+\overline{y,(z)}+\overline{z,(x)} =6 + \frac{6}{9} = 6+\frac{2}{3} = 6+0.(6)=6.(6)$

b) In mod asemanator putem scrie forma generala a unui numar de tipul $\overline{a,b(c)}$  ca fractie ordinale, unde a, b, c sunt cifre in baza 10 :

$\overline{a,b(c)} = \frac{\overline{abc}-\overline{ab}}{90} = \frac{100a+10b+c-10a-b}{90} = \frac{90a+9b+c}{90} = a+\frac{b}{10} +\frac{c}{90}$

Deci,

$\overline{x,y(z)} = x+\frac{y}{10} + \frac{z}{90} \\\\\overline{y,z(x)} = y+\frac{z}{10} + \frac{x}{90}\\ \\\overline{z,x(y)} = z+\frac{x}{10} + \frac{y}{90}$

Daca le adunam :

$suma = x+y+z+\frac{x+y+z}{10} + \frac{x+y+z}{90}$, inlocuim x+y+z cu 6 si obtinem :

$suma = 6+\frac{6}{10} +\frac{6}{90} = 6+0.6+0.0(6)=6.6(6)=6.(6)$

► Nota :

◘ Daca ai uitat transformarea fractiilor din forma zecimala in forma ordinala ai un exemplu aici :
https://brainly.ro/tema/2760639