Question

[nucuoicuapi]

Cho

\(\dfrac{3x-2y}{4}=\dfrac{4y-3z}{9}=\dfrac{2z-4x}{9}\)

Chứng minh rằng:

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

Answer 1

[Mashiro Shiina]

Sửa đề:

$\dfrac{3x-2y}{4}=\dfrac{2z-4x}{9}=\dfrac{4y-3z}{9}$

\(\Leftrightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{27}=\dfrac{2\left(4y-3z\right)}{18}\)

\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{27}=\dfrac{8y-6z}{18}\)

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

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{27}=\dfrac{8y-6z}{18}\)

\(=\dfrac{12x-8y+6z-12x+8y-6z}{16+27+18}=\dfrac{0}{16+27+18}=0\)

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

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