Question

1.Daca x/x+y + y/y+z +z/z+x=1 atunci aratati ca y/x+y +z/y+z + x/z+x apartin nr naturale.

2.Deminstrati ca a1b cu bara deasupra +a2b cu bara deasupra +.....+a9b cu bara deasupra / a5b cu bara deasupra este numar natural patrat perfect.

Answer 1

Sa notam suma ce trebuie demonstrata cu P
$P=\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{z+x}$
Observam ca un ecuatia initiala putem face urmatoarele artificii de calcul
$\frac{x}{x+y}=\frac{x+y-y}{x+y}=1-\frac{y}{x+y}$
$\frac{y}{y+z}=\frac{y+z-z}{y+z}=1-\frac{z}{y+z}$
$\frac{z}{z+x}=\frac{z+x-x}{z+x}=1-\frac{x}{z+x}$
Si acum le putem adunam pe toate 3
$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}=3-(\frac{y}{x+y}+\frac{z}{y+z}+\frac{x}{z+x})=1\Rightarrow 3-P=1\Rightarrow P=3-1=2$ deci este un nr natural
2) Ne uitam la numarator si vedem cat da
    $\overline{a1b}=100a+10*1+b$
    $\overline{a2b}=100a+10*2+b$
    ...........................................................
    $\overline{a9b}=100a+10*9+b$
   Observam ca apar de 9 ori asa ca o sa avem
    $</span>\overline{a1b}+\overline{a2b}+...+\overline{a9b}=9*100a+10*(1+2+..+9)+9b=9*100a+10*\frac{9*10}{2}+9b=9(100a+\frac{100}{2}+b)=9(100a+50+b)$
   Dar stim ca
    $\overline{a5b}=100a+5*10+b=100a+50+b$
Deci cand impartim cele doua obtinem
    $\frac{\overline{a1b}+\overline{a2b}+...+\overline{a9b}}{\overline{a5b}}=\frac{9(100a+50+b)}{<span>100a+50+b</span>}=9=3^{2}$ deci e patrat perfect