Răspunsuri

2014-07-10T13:29:28+03:00
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








2014-07-10T13:37:29+03:00
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

x =  \frac{k}{4} 


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

 \frac{9*k^{2}}{24} = 54


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







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