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

=>x=ak+2bk+ck; y=2ak+bk-ck; z=4ak-4bk+ck

=> \(\frac{a}{x+2y+c}\)=\(\frac{a}{ak+2bk+ck+4bk+2bk-2ck+4ak-4bk+ck}\)=\(\frac{a}{9ak}\)=\(\frac{1}{9k}\)

Tương tự => \(\frac{a}{x+2y+c}\)=\(\frac{b}{2x+y-z}\)=\(\frac{c}{4x-4y+z}\)=\(\frac{1}{9k}\)

\(\text{Ta có: }\)

\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{9a}\left(1\right)\)

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

\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{4x-4y+z}{9c}\left(3\right)\)

\(\text{Từ (1),(2) và (3) ta có:}\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\text{ hay }\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x+4y+z}\left(\text{nghịch đảo lên rồi chia tất cả cho 9}\right)\)

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

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

Ta có: \(A=\frac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b-c}=\frac{a+2y+z}{9a}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

Từ (1), (2), (3) suy ra \(\frac{x+2y+z}{9a}\)\(=\frac{2x+y-z}{9b}\)\(=\frac{4a-4y+z}{9c}\)

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

\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)(vì tất cả các tử và mẫu khác 0)

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

Sửa đề trong bài làm luôn nhé

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

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

\(\Rightarrow\frac{a+2b-c}{x}=\frac{2\left(2a+b+c\right)}{2y}=\frac{4b+c-4a}{z}=\frac{9a}{x+2y-z}\left(1\right)\)

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

\(\Rightarrow\frac{-4\left(a+2b-c\right)}{-4x}=\frac{4\left(2a+b+c\right)}{4y}=\frac{4b+c-4a}{z}=\frac{9c}{-4x+4y+z}\left(3\right)\)

Từ (1), (2), (3) ta có ĐPCM

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

\(\Rightarrow\frac{x}{a+2b-c}=\frac{2y}{4a+2b+c}=\frac{z}{4b+c-4a}=\frac{x+2y-z}{9a}\left(1\right)\)

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

\(\Rightarrow\frac{4x}{4a+8b-4c}=\frac{4y}{8a+4b+4c}=\frac{z}{4b+c-4a}=\frac{4y+z-4a}{9c}\left(3\right)\)

Từi (1),(2),(3) 

còn j giải típ nha

@@@@@@@@@@@@

Ta có: \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x}{2a+4b+2c}=\frac{2y}{4a+2b-2c}\)

\(=\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{\left(a+2b+c\right)+\left(4a+2b-2c\right)+\left(4a-4b+c\right)}=\frac{x+2y+z}{9a}\left(1\right)\)

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

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

Từ (1); (2); (3) \(\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)

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

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

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