Răspuns :
[tex]x * y=x+y-\frac{x y}{3}[/tex]
a) Il vom inlocui pe x cu 1 si pe y cu 3 si obtinem:
[tex]1 * 3=1+3-\frac{1\times3}{3}=1+3-1=3[/tex]
b)
Calculam x*x
[tex]x * x=x+x-\frac{x ^2}{3}=2x-\frac{x ^2}{3}=\frac{6x-x^2}{3} \\\\(\frac{6x-x^2}{3} )*x=\frac{6x-x^2}{3} +x-\frac{\frac{6x-x^2}{3} \times x}{3} =\frac{18x-3x^2+9x-6x^2+x^3}{9} \\\\\frac{18x-3x^2+9x-6x^2+x^3}{9}=\frac{26}{9}[/tex]
Avem acelasi numitor comun, deci vom elimina numitorul si vom obtine:
x³-9x²+27x=26
x³-3x²+27x-26=0
x³-3x²+27x-27+1=0
Avvem urmatoare formula de calcul prescurtat:
(x-y)³=x³-3x²y+3xy²-y³
(x-3)³=x³-9x²+27x-27
Deci x³-3x²+27x-27+1=0
(x-3)³+1=0
(x-3)³=-1
x-3=-1
x=2
c)
Elementul neutru:
x*e=x
[tex]x+e-\frac{x e}{3}=x[/tex]
Aducem la acelasi numitor comun 3
3x+3e-xe=3x
3e-xe=0
e(3-x)=0
e=0
Elementul simetric:
n*n'=e
[tex]n+n'-\frac{nn'}{3}=e\\\\n+n'-\frac{nn'}{3}=0[/tex]
Aducem la acelasi numitor comun 3
3n+3n'-nn'=0
n'(3-n)=-3n |×(-1)
n'(n-3)=3n
[tex]n'=\frac{3n}{n-3}[/tex]
Cum n si n' sunt numere naturale ⇒ n-3 | 3n
n-3 | 3n
n-3 | n-3 |×3
n-3 | 3n
n-3 | 3n-9
Le scadem:
n-3 | 9
n-3={1,-1,3,-3,9,-9}
n-3={4,2,6,0,12,-6}
Dar n si n' ∈N
n={0,4,6,12}
Un exercitiu similar de bac gasesti aici: https://brainly.ro/tema/722385
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