Question

giúp mk nhé

Answer 1

Đặt A=\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)

Ta có:

A=\(\dfrac{x}{a+2b+c}=\dfrac{y}{4a+2b-2c}=\dfrac{z}{4a-4b+c}\)=\(\dfrac{x+y+z}{a+2b+c+4a+2b-2c+4a-4b+c}\)=\(\dfrac{x+2y+z}{9a}\)(1)

A=\(\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)=\(\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}=\dfrac{2x+y-z}{9b}\)(2)

A=\(\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}\)=\(\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\)=\(\dfrac{4x-4y+z}{9c}\)(3)

Từ (1), (2) và (3):

\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)

\(\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)

\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)