Question

[huynhthanhtruc]

cho 3x-2y/2=2z-4x/3=4y-3x/2. chung minh x/2=y/3=z/4

Answer 1

[Vũ Minh Tuấn]

Sửa lại đề là \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}.\)

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

\(\Rightarrow\frac{4.\left(3x-2y\right)}{16}=\frac{3.\left(2z-4x\right)}{9}=\frac{2.\left(4y-3z\right)}{4}.\)

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}\frac{3x-2y}{4}=0\\\frac{2z-4x}{3}=0\\\frac{4y-3z}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-2y=0\\2z-4x=0\\4y-3z=0\end{matrix}\right.\)

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

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