Question

Daca media aritmetica a nr. x,y,z este 6 si ele sunt invers proportionale cu 2,3 si 6 => x= ; y= ; z= .

Answer 1

{x,y,z} i.p. {2,3,6}
2/x= 3/y= 6/z =k=>
x= k/2
y= k/3
z= k/6
(x+y+z)/3=6=> x+y+z= 18
k/2+k/3+k/6= 18=> (3k+2k+k)/6= 18=> 6k= 18*6=> k= 18*6/6=> k= 18
Daca k= 18=> x= k/2= 18/2= 9
y= k/3= 18/3= 6
z= k/6= 18/6= 3
R: x= 9; y= 6; z= 3

Answer 2

(x+y+z):3=6⇒x+y+z=6•3=18
x/¹/₂= y/¹/₃ = z/¹/₆ = x+y+z supra ¹/₂+¹/₃+¹/₆
Amplificam ¹/₂ cu 3 ; ¹/₃ amplificam cu 2.
x+y+z supra ³/₆+²/₆+¹/₆= 18 supra ⁶/₆=18/1= 18
x supra ¹/₂= 18⇒x= 18•¹/₂= 18/2=9
y supra ¹/₃=18⇒y=18/3=6
z supra ¹/₆= 18⇒z=18/6=3