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
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