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

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

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

\(\frac{x+y}{15}=\frac{z+x}{10}=\frac{\left(x+y\right)-\left(z+x\right)}{15-10}=\frac{y-z}{5}\left(1\right)\)

\(\frac{z+x}{10}=\frac{y+z}{6}=\frac{\left(z+x\right)-\left(y+z\right)}{10-6}=\frac{x-y}{4}\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{y-z}{5}=\frac{x-y}{4}\)

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:

Cop mạng ghi nguồn đầy đủ vào nhé!

Ta có:  \(2\left(x+y\right)=3\left(z+x\right)\)

\(\Rightarrow\frac{x+y}{3}=\frac{z+x}{2}\)

\(=\frac{x+y-\left(z+x\right)}{3-2}=y-z\)(tính chất dãy tỉ số bằng nhau)

\(\Rightarrow\frac{x+z}{2}=y-z\)

\(\Rightarrow\frac{x+z}{10}=\frac{y-z}{5}\left(1\right)\)

Lại có:\(5\left(y+z\right)=3\left(x+z\right)\)

\(\Rightarrow\frac{y+z}{3}=\frac{x+z}{5}\)

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

\(\Rightarrow\frac{x+z}{5}=\frac{x-y}{2}\)

\(\Rightarrow\frac{x+z}{10}=\frac{x-y}{4}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{x-y}{4}=\frac{y-z}{5}\left(đpcm\right)\)

Câu hỏi của Nguyễn Quang Tùng - Toán lớp 7 - Học toán với OnlineMath

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

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

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

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

Ta có : \(\frac{x+y}{3}=\frac{z+x}{2}=\frac{x+y-z-x}{3-2}=y-z\)

\(\frac{y+z}{3}=\frac{z+x}{5}=\frac{-y-z+z+x}{-3+5}=\frac{x-y}{2}\)

\(\frac{x-y}{2}=\frac{z+x}{5}\)\(\Rightarrow x-y=\frac{2.\left(z+x\right)}{5}\)\(\Rightarrow\frac{x-y}{4}=\frac{z+x}{10}\)( 1 )

\(\frac{z+x}{2}=y-z\)\(\Rightarrow\frac{y-z}{5}=\frac{z+x}{10}\)                                               ( 2 )

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{x-y}{4}=\frac{y-z}{5}\)

Vì 5(y+z)=3(z+x) =>(x+z)/5=(y+z)/3=(x+z-y-z)/(5-3) = (x-y)/2

Ap dung tinh chat day ti so bang nhau ta co : 

Do đó (x+z)/5 = (x-y)/2 \(\Leftrightarrow\) (x+z)/10=(x-y)/4 (1)  

Ta lại có: 2(x+y)=3(z+x) \(\Rightarrow\) (x+z)/2=(x+y)/3=(x+z-x-y)/(2-3)=y-z

Ap dung tinh chat day ti so bang nhau ta co : 

Do đó (x+z)/2 = y-z \(\Leftrightarrow\) (x+z)/10=(y-z)/5 (2)  

Từ (1) và (2) suy ra (x-y)/4=(y-z)/5

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

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

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

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

Vậy : ...

\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=k\Rightarrow\hept{\begin{cases}x+y=15k\\y+z=6k\\z+x=10k\end{cases}\Rightarrow\hept{\begin{cases}x-y=4k\\y-z=5k\end{cases}\Rightarrow}\frac{x-y}{4}=\frac{y-z}{5}}\)