Question

Chứng minh rằng nếu : 

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

Answer 1

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

 

 

Answer 2

cảm ơn bn nhiều

Answer 3

thanks

Answer 4

cảm ơn bạn nhé

Answer 5

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

Answer 6

thank you các bạn nhìu ! mãi yêu mọi người =))))

Answer 7

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

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

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

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

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

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