Question

Cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\). Chứng minh rằng \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)

Answer 1

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

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

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

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y-z}{9c}\)

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

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