Question

Determinati numerele x , y , z stiind ca sunt direct proportionalecu 0.25 ; 0.(3) ,si 0.5 si ca xy+xz+yz =3xyz

Answer 1

x=0,25m=m/4
y=0,(3)m=m/3
z=0,5m=m/2, unde m=N*
xy+xz+yz =3xyz
(m^2)/12+(m^2)/8+(m^2)/6=(3m^3)/24
Inmultim cu 24:
2(m^2)+3(m^2)+4(m^2)=3m^3
3m^3-9m^2=0
3m^2(m-3)=0
3m^2>0,=>m-3=0, m=3
x=3/4
y=1
z=3/2

Answer 2

[tex] \frac{x}{0,25} =\frac{y}{0,(3)} = \frac{z}{0,5}=k \\ x=0,25k= \frac{k}{4} \\ y=0,(3)k= \frac{k}{3} \\ z=0,5k= \frac{k}{2 } \\ \frac{k^2}{12} + \frac{k^2}{8} + \frac{k^2}{6} =3 \frac{k^3}{24} \\ k^2( \frac{1}{12} + \frac{1}{8}+ \frac{1}{6})= \frac{k^3}{8} \\ \frac{2+3+4}{24}= \frac{3k}{24} \\ 3k=9 \\ k=3 \\ x=0,25*3=0,75;y=0,(3)*3=1;z=0,5*3=1.5[/tex]