[polsk133]

$\frac{(logx)^2}{logylogz} + \frac{(logy)^2}{logxlogz}+\frac{(logz)^2}{logxlogy} -3 =0$

คูณ$ logxlogylogz$ ตลอด

ได้$ (logx)^3+(logy)^3+(logz)^3 - 3 logxlogylogz=0$

$(logx+logy+logz)((logx)^2+(logy)^2+(logz)^2-logxlogy-logylogz-logzlogx)=0$

CASE 1
$logx+logy+logz=0$

$logxyz=0$

$xyz=1$

CASE2

$(logx)^2+(logy)^2+(logz)^2-logxlogy-logylogz-logzlogx=0$

$(logx-logy)^2+(logy-logz)^2+(logz-logx)^2=0$

$log x=logy=logz $

$x=y=z >0$ แล้วทำไงต่อหละเนี่ย

ตอบ $R^{+}$ ละกันครับ