Question

Aflati numerele naturale x,y,z stiind ca sunt respectiv proportionale cu 0,25; 0,(3); 0,5 si ca xy +xz +yz =54.

Answer 1

0,25=1/4
0,(3)=1/3
0,5=1/2
"x,y,z stiind ca sunt respectiv proportionale cu 0,25; 0,(3); 0,5" inseamna ca" exista k nr nat astfel incat:
x/0,25=y/0,(3)=z/0,5=k, adica
x/(1/4)=y/(1/3)=z/(1/2)=k si facand calculele:
4x=3y=2z=k, deci:
x=k/4, y=k/3 si z=k/2 si inlocuim in relatia:

xy +xz +yz =54
k*k/(3*4)+k*k/(2*4)+k*k/(3*2)=54 Aducem la acelasi numitor:
k*k(2+3+4)/(2*3*4)=9*6
k*k*9/(2*3*4)=9*6 impartim prin 9 si unmultim cu 2*3*4:
k*k=6*6*4=144 deci k=12, de unde x=3, y=4, z=6

Answer 2

0,25=$\frac{25}{100} = \frac{1}{4}$

$0,(3) = \frac{3}{9} = \frac{1}{3}$

$0,5= \frac{5}{10} = \frac{1}{2}$

$\frac{x}{ \frac{1}{4} } = \frac{y}{ \frac{1}{3}} = \frac{z}{ \frac{1}{2} }$ = k

$\frac{x}{1} : \frac{1}{4} = \frac{y}{1} : \frac{1}{3} = \frac{z}{1} : \frac{1}{2} = k$

4x = 3y = 2z = k

[tex]x = \frac{k}{4} [/tex]

$y = \frac{k}{3}$$\frac{k}{4}$

$z = \frac{k}{2}$

$\frac{k}{4}$ * $\frac{k}{3}$ + $\frac{k}{4}$ * $\frac{k}{2}$ + $\frac{k}{3}$ * $\frac{k}{2}$ = 54

$\frac{k^{2} }{12} + \frac{k^{2} }{8} + \frac{k^{2} }{6} = 54$

aducem la acelasi numitor

$\frac{2*k^{2} }{24} + \frac{3*k^{2} }{24} + \frac{4*k^{2} }{24} = 54$

[tex] \frac{9*k^{2}}{24} = 54 [/tex]

$9* k^{2} = 24*54$

$9* k^{2} = 1296$        /:9

$k^{2} =144$

k=12

x=12:4=3

y=12:3=4

z=12:2=6