Từ đẳng thức đã cho ta có
$\dfrac{3x-2y}{4} = \dfrac{2z-4x}{3} = \dfrac{4y - 3z}{2}$
Ta có
$\dfrac{3x-2y}{4} =\dfrac{4y - 3z}{2} = \dfrac{6x-4y}{8} = \dfrac{6x-4y+4y-3z}{8+2} = \dfrac{6x-3z}{10}$
Lại có
$\dfrac{2z-4x}{3} = \dfrac{3z-6x}{\frac{9}{2}}$
Vậy ta có
$\dfrac{3z-6x}{\frac{9}{2}} = \dfrac{6x-3z}{10} = \dfrac{3z-6x+6x-3z}{10 + \frac{9}{2}} = \dfrac{0}{\frac{29}{2}} = 0$
Vậy ta suy ra
$\dfrac{3x-2y}{4} = \dfrac{2z-4x}{3} = \dfrac{4y - 3z}{2} = 0$
hay
$3x-2y = 2z-4x = 4y-3z = 0$
Do $3x - 2y = 0$ suy ra $\dfrac{x}{2} = \dfrac{y}{3}$
Do $2z-4x = 0$ suy ra $\dfrac{x}{2} = \dfrac{z}{4}$
Do $4y - 3z = 0$ suy ra $\dfrac{y}{3} = \dfrac{z}{4}$
Vậy
$\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$
