Ta có: $\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}$
\[\Rightarrow \frac{a+2b+c}{x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}\]
\[\Rightarrow \frac{a+2b+c}{x}=\frac{2(2a+b-c)}{2y}=\frac{4a-4b+c}{z}
=\frac{a+2b+c+4a+2b-2c+4a-4b+c}{x+2y+z}=\frac{9a}{x+2y+z}\]
\[\Rightarrow \frac{2(a+2b+c)}{2x}=\frac{2a+b-c}{y}=\frac{2a-4b+c}{z}
=\frac{4a+4b+2c+2a+b-c-4a+4b-c}{2x+y-z}=\frac{9b}{2x+y-z}\]
\[\Rightarrow \frac{4(a+2b+c)}{4x}=\frac{4(2a+b-c)}{4y}=\frac{4a-4b+c}{z}
=\frac{4a+8b+4c-8a-4b+4c+4a-4b+c}{4x-4y+z}=\frac{9c}{4x-4y+z}\]
\[\Rightarrow \frac{9a}{x+2y+z}=\frac{9b}{2x+y-z}=\frac{9c}{4x-4y+z}\]
\[\Rightarrow \frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\]
(ĐPCM)