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 : \(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR : nếu 2 . ( x + y ) = 5 . ( y + z ) = 3 . ( z + x ) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR nếu 2(x+y)=5(y+z)=3(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR nếu 2(x+y)=5(y+z)=3(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR: Nếu 2.(x+y) = 5.(y+z)=3.(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR:2(x+y)=5(y+z)=3(z+x) thì\(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR 2(x+y)=5(y+z) =3(z+x) thì\(\frac{x-y}{4}=\frac{y-z}{5}\)
CMR:2(x+y)=5(y+z)=3(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
Chứng minh rằng: Nếu 2(x+y) = 5(y+z) = 3(z+x) thì \(\frac{x-y}{4}=\frac{y-z}{5}\)
Nếu 2(x+y)=5(y+z)=3(z+x) thì \(\frac{x-y}{4}\)=\(\frac{y-z}{5}\)
