Ta có:
`(3x - 2y) : 4 = (2z - 4x) : 3 = (4y - 3z) : 2`
⇒ $\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}$
⇒ $\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}$
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
$\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}$
$=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}$ = 0
⇒ $\dfrac{12x-8y}{16} = 0\Rightarrow 12x-8y = 0 \Rightarrow 12x = 8y $
$\Rightarrow \dfrac{x}{8}=\dfrac{y}{12} \Rightarrow \dfrac{x}{2}=\dfrac{y}{3}$ (1)
$\dfrac{6z-12x}{9} = 0 \Rightarrow 6z - 12x = 0 \Rightarrow 6z = 12x$
$\Rightarrow \dfrac{x}{6}=\dfrac{z}{12} \Rightarrow \dfrac{x}{2}=\dfrac{z}{4}$ (2)
Từ (1) và (2)
⇒ $\dfrac{x}{2} =\dfrac{y}{3}=\dfrac{z}{4}$
$\Rightarrow$ `x : 2 = y : 3 = z : 4` (đpcm)
