Ta có:
$2(x+y) = 5(y+z) = 3(z+x)$
$+) \quad 2(x+y) = 3(z+x)$
$\to \dfrac{x+y}{3} = \dfrac{z+x}{2}$
$\to \dfrac{x+y}{3} = \dfrac{z+x}{2} = \dfrac{x +y - z -x}{3 - 2} = \dfrac{y-z}{1} = y -z$
$\to \dfrac{z+x}{2} = y - z$
$\to \dfrac{z+x}{10} = \dfrac{y-z}{5}\qquad (1)$
$+) \quad 5(y+z) = 3(z+x)$
$\to \dfrac{y+z}{3} = \dfrac{z+x}{5}$
$\to \dfrac{y+z}{3} = \dfrac{z+x}{5} = \dfrac{z+x- y- z}{5 - 3} = \dfrac{x-y}{2}$
$\to \dfrac{z+x}{5} = \dfrac{x-y}{2}$
$\to \dfrac{z+x}{10} = \dfrac{x-y}{4}\qquad (2)$
$(1)(2)\to \dfrac{x-y}{4} = \dfrac{y-z}{5}$
