Ta có \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4z-3z}{2}\)
⇒ \(\frac{12x-8y}{16}=\frac{6z-12x}{6}=\frac{8y-6z}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có
\(\frac{12x-8y}{16}=\frac{6z-12x}{6}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{4+6+16}=0\)
⇒ \(\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\z=2x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6x=4y\\3z=6x\end{matrix}\right.\)
\(\Rightarrow6x=4y=3z\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
