@Người dùng ẩn 552547: 06:36 09/06/2016

Biết rằngbz−cya = cx−azb = ay−bxc

Hãy chứng minh x:y:z = a:b:c

Ta có:

$\dfrac{bz-cy}{a}$ = $\dfrac{cx-az}{b}$ = $\dfrac{ay-bx}{c}$ = $\dfrac{bxz-cxy}{ax}$ = $\dfrac{cxy-ayz}{by}$ = $\dfrac{ayz-bxz}{cz}$ = $\dfrac{0}{ax+by+cz}$ = 0

$\begin{array}{l}
 \Rightarrow bz = cy \Rightarrow \dfrac{z}{c} = \dfrac{y}{b}\left( 1 \right)\\
 \Rightarrow cx = az \Rightarrow \dfrac{x}{a} = \dfrac{z}{c}\left( 2 \right)\\
 \Rightarrow ay = bx \Rightarrow \dfrac{y}{b} = \dfrac{x}{a}\left( 3 \right)
\end{array}$

Từ (1),(2),(3) $ \Rightarrow \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ hay x:y:z = a:b:c