=> \(\frac{\text{2(x+y)}}{30}\)=\(\frac{\text{5(y+z)}}{30}\)=\(\frac{\text{3(z+x)}}{30}\)

=> \(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)

Theo t/c dãy tỉ số bằng nhau có:

\(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)=\(\frac{\left(z+x\right)-\left(y+z\right)}{10-6}\)=\(\frac{x-y}{4}\)*

\(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)=\(\frac{\left(x+y\right)-\left(z+x\right)}{15-10}\)=\(\frac{y-z}{5}\)**

Từ * và ** => \(\frac{x-y}{4}\)=\(\frac{y-z}{5}\)(đpcm)

K cần t i c k 

\(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\Leftrightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{y+z}{6}=\frac{z+x}{10}=\frac{\left(z+x\right)-\left(y+z\right)}{10-6}=\frac{x-y}{4}\)

\(\frac{x+y}{15}=\frac{z+x}{10}=\frac{\left(x+y\right)-\left(z+x\right)}{15-10}=\frac{y-z}{5}\)

Suy ra đpcm. 

Vì 5(y+z) = 3(x+z)

Suy ra (x+z) / 5 = (y+z) / 3 = (x+z-y-z) / 5-3 = (x-y) / 2

Suy ra (x+z) / 5 = (x-y) / 2 tương đương (x+z) / 10 = (x-y) / 4                               (1)

2(x+y) = 3(x+z)

Suy ra (x+z) / 2 = (x+y) / 3 = (x+z-x-y) / 2-3 = y-z

(x+z) / 2 = y-z

Tương đương (x+z) / 10 = (y-z) / 5                                                                      (2)

Từ (1) và (2) suy ra:

Lời giải:

$2(x+y)=5(y+z)=3(z+x)$

$\Rightarrow \frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}$

Đặt $\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=t$

$\Rightarrow x+y=15t; y+z=6t; z+x=10t$

$\Rightarrow 2(x+y+z)=x+y+y+z+z+x=15t+6t+10t=31t$

$\Rightarrow x+y+z=15,5t$

$z=(x+y+z)-(x+y)=15,5t-15t=0,5t$

$x=(x+y+z)-(y+z)=15,5t-6t=9,5t$

$y=(x+y+z)-(x+z)=15,5t-10t=5,5t$

Suy ra:

$\frac{x-y}{4}=\frac{9,5t-5,5t}{4}=t$

$\frac{y-z}{5}=\frac{5,5t-0,5t}{5}=t$

$\Rightarrow \frac{x-y}{4}=\frac{y-z}{5}$