Question

[0o0^^^Nhi^^^0o0]

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

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

Answer 1

[Nguyễn Thị Huyền Trang]

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

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

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

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

Theo t/c dãy tỉ số bằng nhau, ta có:

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

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

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

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

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

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