Răspuns
[tex]x = +/- \frac{1}{4} , y = +/-\frac{3}{8} , z = +/- \frac{1}{2}[/tex]
Explicație pas cu pas:
x×6 = y×4 = z×3 = k, de unde x=k/6 y=k/4 z=k/3 (1)
[tex]\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 24[/tex] ⇔ [tex]\frac{x+y+z}{xyz} = 24[/tex] (2)
In relatia (2) inlocuim pe x,y si z conform egalitatilor din (1). Obtinem
[tex]\frac{\frac{k}{3}+\frac{k}{6}+\frac{k}{4} }{\frac{k}{3}*\frac{k}{6}*\frac{k}{4} } = 24[/tex] ⇔ [tex]\frac{\frac{4k+2k+3k}{12} }{\frac{k^{3} }{72} } = 24[/tex] ⇔ [tex]\frac{9k}{12} * \frac{72}{k^{3} } = 24[/tex] ⇔ [tex]\frac{9*6}{k^{2} } = 24[/tex] ⇔
⇔ 24×k² = 9×6 ⇔ 4k² = 9 de unde k = ± [tex]\frac{3}{2}[/tex]
Solutia 1: k=3/2
x=k/6 = 3/12 = 1/4
y=k/4 = 3/8
z=k/3 = 3/6 = 1/2
Solutia 2: k=-3/2
x=k/6 = -3/12 = -1/4
y=k/4 = -3/8
z=k/3 = -3/6 = -1/2
