\(\frac{3}{x}=\frac{x}{12}\Leftrightarrow x^2=3.12=36\Leftrightarrow x^2=6^2=\left(-6\right)^2\Leftrightarrow x\in\left\{-6;6\right\}\)
+) Với x = -6:
\(-\frac{6}{12}=-\frac{1}{2}=\frac{y+1}{4}\Leftrightarrow2.\left(y+1\right)=-4\Leftrightarrow y+1=-2\Leftrightarrow y=-3\)
\(-\frac{6}{12}=-\frac{1}{2}=\frac{z^2-1}{16}\Leftrightarrow2.\left(z^2-1\right)=-16\Leftrightarrow z^2-1=-8\Leftrightarrow z^2=-7\left(\text{vô lí}\right)\)
=> Không có x , y , z thỏa mãn.
+) Với x = 6:
\(\frac{6}{12}=\frac{y+1}{4}\Leftrightarrow12.\left(y+1\right)=24\Leftrightarrow y+1=2\Leftrightarrow y=1\)
\(\frac{6}{12}=\frac{z^2-1}{16}\Leftrightarrow12.\left(z^2-1\right)=96\Leftrightarrow z^2-1=8\Leftrightarrow z^2=9\Leftrightarrow z\in\left\{-3;3\right\}\)
Vậy các cặp (x;y;z) thỏa là: (6;1;-3); (6;1;3).
