Câu hỏi của Luchia

Question

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

Chứng minh x,y,z tỉ lệ với 2,3,4

Nguyễn Thanh Hằng · Accepted Answer

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

$\Leftrightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}$

$\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}$

Áp dụng t,c dãy tỉ số bằng nhau ta có :

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

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-8y}{16}=0\\dfrac{2z-4x}{3}=0\\dfrac{4y-3z}{2}=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}12x-8y=0\2x-4z=0\4y-3z=0\end{matrix}\right.$

$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\dfrac{y}{3}=\dfrac{z}{4}\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.$

$\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)$Câu trả lời: $\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}$