Giải thích các bước giải:
Ta có:
$\dfrac{2x-3y}{5}=\dfrac{3y-4z}{6}=\dfrac{4z-2x}{7}=\dfrac{(2x-3y)+(3y-4z)+(4z-2x)}{5+6+7}=\dfrac{0}{18}=0$
$\to 2x-3y=3y-4z=4z-2x=0$
$\to 2x=3y, 3y=4z, 4z=2x$
$\to 2x=3y=4z$
$\to \dfrac{2x}{12}=\dfrac{3y}{12}=\dfrac{4z}{12}$
$\to \dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}$