ĐKXĐ: \(z\ne0\)
Ta có: \(3x=4y=6z\)
\(\Leftrightarrow\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{6}}\)
Đặt \(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{6}}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{k}{3}\\y=\frac{k}{4}\\z=\frac{k}{6}\end{matrix}\right.\)
Ta có: \(\frac{xy}{z}=-18\)
\(\Leftrightarrow xy=-18z\)
\(\Leftrightarrow\frac{k}{3}\cdot\frac{k}{4}=-18\cdot\frac{k}{6}\)
\(\Leftrightarrow\frac{k^2}{12}=\frac{-18k}{6}\)
\(\Leftrightarrow\frac{k^2}{12}=-3k\)
\(\Leftrightarrow k^2=-3k\cdot12=-36k\)
\(\Leftrightarrow k^2+36k=0\)
\(\Leftrightarrow k\left(k+36\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}k=0\\k+36=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k=0\\k=-36\end{matrix}\right.\)
Trường hợp 1: k=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{k}{3}=\frac{0}{3}=0\\y=\frac{k}{4}=\frac{0}{4}=0\\z=\frac{k}{6}=\frac{0}{6}=0\left(loại\right)\end{matrix}\right.\)
Trường hợp 2: k=-36
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{k}{3}=\frac{-36}{3}=-12\\y=\frac{k}{4}=\frac{-36}{4}=-9\\z=\frac{k}{6}=\frac{-36}{6}=-6\left(nhận\right)\end{matrix}\right.\)
Vậy: (x,y,z)=(-12;-9;-6)
