Răspuns :
x + y + z = 120
{x ;y ;z} d.p. cu { 3;5;16 }
=>
[tex] \frac{x}{3} = \frac{y}{5} = \frac{z}{16} = k \\ = > x = 3k \: \: y = 5k \: \: z = 16k[/tex]
Se inlocuiesc in suma:
3k + 5k + 16k = 120
24k = 120
k= 5 => x = 15 ; y =25 ; z = 80
{x ;y ;z} d.p. cu { 3;5;16 }
=>
[tex] \frac{x}{3} = \frac{y}{5} = \frac{z}{16} = k \\ = > x = 3k \: \: y = 5k \: \: z = 16k[/tex]
Se inlocuiesc in suma:
3k + 5k + 16k = 120
24k = 120
k= 5 => x = 15 ; y =25 ; z = 80
{ a, b, c } d,p { 3,5,16 }
--------------------------/-------
a + b + c = 120
---------------------------//-----
a / 3 = b / 5 = c / 16 = K
a / 3 = K => a = 3 K
b / 5 = K => b = 5 K
c / 16 = K => c = 16 K
3 K + 5 K + 16 K = 120
8 K + 16 K = 120
24 K = 120
K = 120 : 24
K = 5
a = 5 x 3 = 15
b = 5 x 5 = 25
c = 5 x 16 = 80
