Question

[Nguyễn Thanh Hằng]

Tìm x, y, z biết: \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\) và x + y + z = 18

Answer 1

[Trúc Giang]

Ta có: \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)

\(\Rightarrow\frac{4\left(3x-2y\right)}{4.4}=\frac{3\left(2z-4x\right)}{3.3}=\frac{2\left(4y-3z\right)}{2.2}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

\(\left\{{}\begin{matrix}\frac{12x-8y}{16}=0\Rightarrow12x-8y=0\Rightarrow12x=8y\\\frac{6z-12x}{9}=0\Rightarrow6z-12x=0\Rightarrow6z=12x\end{matrix}\right.\)

\(\Rightarrow12x=8y=6z\)

\(\Rightarrow\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=\frac{x+y+z}{\frac{1}{12}+\frac{1}{8}+\frac{1}{6}}=\frac{18}{\frac{3}{8}}=18.\frac{8}{3}=48\)

\(\left\{{}\begin{matrix}\frac{x}{\frac{1}{12}}=48\Rightarrow x=48.\frac{1}{12}=4\\\frac{y}{\frac{1}{8}}=48\Rightarrow y=48.\frac{1}{8}=6\\\frac{z}{\frac{1}{6}}=48\Rightarrow z=48.\frac{1}{6}=8\end{matrix}\right.\)