Giải thích các bước giải:
Ta có:
$\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=k$
$\to k=\dfrac{x+2y+z}{(a+2b+c)+2(2a+b-c)+(4a-4b+c)}=\dfrac{2x+y-z}{2(a+2b+c)+(2a+b-c)-(4a-4b+c)} =\dfrac{4x-4y+z}{4(a+2b+c)-4(2a+b-c)+(4a-4b+c)}$
$\to k=\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b} =\dfrac{4x-4y+z}{9c}$
$\to \dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b} =\dfrac{4x-4y+z}{9c}$
$\to \dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b} =\dfrac{4x-4y+z}{c}$
$\to \dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z} =\dfrac{c}{4x-4y+z}$
$\to đpcm$
