Question

\(2.\left(x-y\right)=5.\left(y+z\right)=3.\left(x+z\right)\)

CM : \(\frac{x-y}{4}=\frac{y-z}{5}\)

Answer 1

Theo đề ta có:

\(2\left(x+y\right)=5\left(y+z\right)=3\left(x+z\right)\)

\(\Rightarrow\)\(\frac{2\left(x+y\right)}{30}=\frac{5\left(y+z\right)}{30}=\frac{3\left(x+z\right)}{30}\)

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

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

* \(\frac{y+z}{6}=\frac{x+z}{10}=\frac{\left(x+z\right)-\left(y+z\right)}{10-6}=\frac{x+z-y-z}{4}=\frac{x-y}{4}\)  \(\left(1\right)\)

* \(\frac{x+y}{15}=\frac{x+z}{10}=\frac{x+y-\left(x+z\right)}{15-10}=\frac{x+y-x-z}{5}=\frac{y-z}{5}\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)ta có:

\(\frac{x-y}{4}=\frac{y-z}{5}\)