ta có: \(5.\left(y+z\right)=3.\left(z+x\right)\)
\(\Rightarrow\frac{z+x}{5}=\frac{y+z}{3}=\frac{z+x-y-z}{5-3}=\frac{x-y}{2}\)
\(\Rightarrow\frac{z+x}{5}=\frac{x-y}{2}\Rightarrow\frac{1}{2}.\frac{z+x}{5}=\frac{1}{2}.\frac{x-y}{2}=\frac{z+x}{10}=\frac{x-y}{4}\) (1)
ta có: \(2.\left(x+y\right)=3.\left(z+x\right)\)
\(\Rightarrow\frac{x+y}{3}=\frac{z+x}{2}=\frac{x+y-z-x}{3-2}=\frac{y-z}{1}=y-z\)
\(\Rightarrow\frac{z+x}{2}=y-z\Rightarrow\frac{1}{5}.\frac{z+x}{2}=\frac{1}{5}.\left(y-z\right)\Rightarrow\frac{z+x}{10}=\frac{y-z}{5}\)(2)
Từ (1);(2) \(\Rightarrow\frac{x-y}{4}=\frac{y-z}{5}\left(=\frac{z+x}{10}\right)\) ( đ p c m)
Ta có: \(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\)
\(\Rightarrow\frac{2\left(x+y\right)}{30}=\frac{5\left(y+z\right)}{30}=\frac{3\left(z+x\right)}{30}\)
\(\Rightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=\frac{x+y-\left(z+x\right)}{15-10}=\frac{z+x-\left(y+z\right)}{10-6}\)
\(\Rightarrow\frac{x-y}{4}=\frac{y-z}{5}\)
