\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{9a}\)(1)
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x+y-z}{9b}\)(2)
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{4x-4y+z}{9c}\)(3)
Từ (1), (2), (3) => \(\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)hay \(\frac{a}{x+2y+z}=\frac{b}{2z+y-z}=\frac{c}{4x-4y+z}\)(vì cùng = 9)
sao\(\frac{x}{a=2b=c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{9a}\)
thank you các bạn nhìu ! mãi yêu mọi người =))))
Từ \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
\(\Rightarrow\frac{a+2b+c}{x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}\)
\(\Rightarrow\frac{a+2b-c}{x}=\frac{2\left(2a+b-c\right)}{2y}=\frac{4a-4b+c}{z}=\frac{a}{x+2y+z}\left(1\right)\)
\(\frac{2\left(a+2b+c\right)}{2x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}=\frac{b}{2x+y+z}\left(2\right)\)
\(\frac{4\left(a+2b+c\right)}{4x}=\frac{4\left(2a+b-c\right)}{4y}=\frac{4a+4b-c}{z}=\frac{c}{4x-4y+z}\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)