Question

cho \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

cmr: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Answer 1

Theo đề ta có:

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

=> \(4.\dfrac{3x-2y}{4}=3.\dfrac{2z-4x}{3}=2.\dfrac{4y-3z}{2}\)

=> \(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

=> \(\dfrac{12x-8y}{16}+\dfrac{6z-12x}{9}+\dfrac{8y-6z}{4}=\dfrac{0}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12x=8y=6z\\\end{matrix}\right.\)

=> \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)( MSC: 24)

=> \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)(đpcm)

Answer 2

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{12x-8x+6x-12x+8y-6z}{16+9+4}\\ =0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)