Giải thích các bước giải:
Ta có:
$\dfrac{1.5x-y}{2}=\dfrac{9z-18x}{13.5}=\dfrac{4y-3z}{2}$
$\to \dfrac{4(1.5x-y)}{4\cdot 2}=\dfrac{(9z-18x):3}{13.5:3}=\dfrac{4y-3z}{2}$
$\to \dfrac{6x-4y}{8}=\dfrac{3z-6x}{4.5}=\dfrac{4y-3z}{2}=\dfrac{6x-4y+3z-6x+4y-3z}{8+4.5+2}=\dfrac{0}{14.5}=0$
$\to 6x-4y=3z-6x=4y-3z=0$
$\to 6x=4y=3z$
$\to\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}$
$\to\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$
