Question

Va rog foarte mult
Sa se afle x,y,z € Q stiind ca sunt invers proportionale cu numerele 0,8 , 0,6 , 0,4
si:
1) x la 2 + y la 2 + z la 2 = 1525
2) 1/x + 1/y + 1/z = 0,25

Answer 1

0,8x = 0,6y = 0,4z = k, adica x*8/10 = y*6/10 = z*4/10 = k, de unde:
x = k:8/10 = k * 10/8 = k*5/4
y = k:6/10 = k*10/6 = k*5/3
z = k:4/10 = k*10/4 = k*5/2

1)x^2 + y^2 + z^2 = 1525
   k^2 * 25/16 + k^2*25/9 + k^2*25/4 = 1525
   k^2(25/16+25/9+25/4) = 1525
   k^2(225/144 + 400/144 + 900/144) = 1525
   k^2 * 1525/144 = 1525
   k^2 = 1525 : 1525/144 = 1525 x 144/1525 = 144
   k = rad144
   k = 12
  Atunci: x = k*5/4 = 12*5/4 = 15
               y = k*5/3 = 12*5/3 = 20
               z = k*5/2 = 12*5/2 = 30

2) 1/x + 1/y + 1/z = 0,25 = 25/10 = 1/4
    1/(5k/4) + 1/(5k/3) + 1/(5k/2) = 1/4
    4/5k  + 3/5k + 2/5k = 1/4
    9/5k = 1/4
    5k = 36
    k = 36/5
    Atunci: x = 5k/4 = (5*36/5)/4 = 36/4 = 9
                y = 5k/3 = (5*36/5)/3 = 36/3 = 12
                z = 5k/2 = (5*36/5)/2 = 36/2 = 18