Question

[Hạ Hy]

Cho \(\dfrac{2x-4y}{3}\) = \(\dfrac{4z-3x}{2}\) = \(\dfrac{3y-2z}{4}\) . Tìm x, y, z biết 2x - y + z = 27

Answer 1

[Nam Nguyễn]

Giải:

Ta có:

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

\(\Rightarrow\dfrac{3\left(2x-4y\right)}{3^2}=\dfrac{2\left(4z-3x\right)}{2^2}=\dfrac{4\left(3y-2z\right)}{4^2}.\)

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

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

\(=\dfrac{\left(6x-6x\right)+\left(8z-8z\right)+\left(12y-12y\right)}{19}=0.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}.\\4z=3x\Rightarrow\dfrac{z}{3}=\dfrac{x}{4}.\\3y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{3}.\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}_{\left(1\right)}\) và \(2x-y+z=27_{\left(2\right)}.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ số bằng nhau có:

\(\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3.\)

Từ đó: \(\left\{{}\begin{matrix}2x=3.8=24\Rightarrow x=12.\\y=3.2=6.\\z=3.3=9.\end{matrix}\right.\)

Vậy.....

Answer 2

[Ngô Tấn Đạt]

\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{6x-12y+8z-6x+12y-8z}{9+4+16}=\dfrac{0}{29}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)