Question

Ofer 20 de puncte, va rog ajutati-ma.

9) Aflati numerele naturale nenule x,y,z stiind ca ele sunt direct proportionale cu primele trei numere naturale prime si xy+ yz - 2xz = z

Answer 1

Primele trei nr. naturale prime:
2:3:5=>{x,y,z}×d.p{2,3,5}=>x/2=y/3=z/5=k=x=2k,y=3k,z=5k

Le inlocuim in a doua relatie
2k×3k+3k×5k+2k×5k=5k

6k^2+15^2+10^2=5k<=>31k^2=5k

Impartim relatia cu k:

31k=5=>5/31=>x=10/31,y=15/31 z=25/31

sau:
x/2=y/3=z/5=k=>x=2k,y=3k si z=5k=>6k^2+q5k^2-20k^2=5k=>k^2=5k k^2-5k=0=>k(k-5)=0=>k1=0 si k2=5 pt k =5=>x=10;y=15 si z=25

Sper ca te-am ajutat coroana?

Answer 2

x/2 = y/3 = z/5 ⇒ z=5y/3,  
xy+ yz - 2xz = z
xy +z(y-2x-1)=0
xy+5y/3   (y-2x-1)=0
dar y=3x/2
3x^2 / 2 + 5*3x/(2*3) (3x/2 - 2x - 1)=0 I *2/x
3x  +  5/2 (3x -4x -2)=0
6x + 15x -20x -10) = 0
x=10
y=3*10/2 =15
z=5*15/3 = 25
Verificare:
10/2=15/3=25/5=5 OK: