Ta có: \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)
\(\Rightarrow12x-8y=6z-12x=8y-6z=0\)
\(\Rightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x=2y\\z=2x\\4y=3z\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3},\dfrac{z}{2}=x,\dfrac{y}{3}=\dfrac{z}{4}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3},\dfrac{z}{4}=\dfrac{x}{2},\dfrac{y}{3}=\dfrac{z}{4}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) (đpcm)
