Ta có:
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2y}{4a+2b-2c}=\frac{x+2y+z}{9a}=\frac{1}{9}.\frac{x+2y+z}{a}\)(1)
\(=\frac{2x}{2a+4b+c}=\frac{2x+y-z}{9b}=\frac{1}{9}.\frac{2x+y-z}{b}\) (2)
\(=\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{4x-4y+z}{9c}=\frac{1}{9}.\frac{4x-4y+z}{c}\) (3)
Từ (1), (2) và (3) => \(\frac{1}{9}.\frac{x+2y+z}{a}=\frac{1}{9}.\frac{2x+y-z}{b}=\frac{1}{9}.\frac{4x-4y+z}{c}\)
=> \(\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}\)
=> \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
