Giải thích các bước giải:
Ta có:
$\dfrac{2by-3az}{x}=\dfrac{3cz-bx}{2y}=\dfrac{ax-2cy}{3z}$
$\to \dfrac{(2by-3az)x}{x^2}=\dfrac{(3cz-bx)y}{2y^2}=\dfrac{(ax-2cy)z}{3z^2}$
$\to \dfrac{2bxy-3azx}{x^2}=\dfrac{3cyz-bxy}{2y^2}=\dfrac{azx-2cyz}{3z^2}=\dfrac{2bxy-3azx+3cyz-bxy+azx-2cyz}{x^2+y^2+z^2}$
$\to \dfrac{2bxy-3azx}{x^2}=\dfrac{3cyz-bxy}{2y^2}=\dfrac{azx-2cyz}{3z^2}$
$\to \dfrac{2bxy-3azx}{x^2}=\dfrac{6cyz-2bxy}{6y^2}=\dfrac{3azx-6cyz}{9z^2}=\dfrac{2bxy-3azx+6cyz-2bxy+3azx-6cyz}{x^2+6y^2+9z^2}$
$\to \dfrac{2bxy-3azx}{x^2}=\dfrac{6cyz-2bxy}{6y^2}=\dfrac{3azx-6cyz}{9z^2}=\dfrac{0}{x^2+6y^2+9z^2}$
$\to \dfrac{2bxy-3azx}{x^2}=\dfrac{6cyz-2bxy}{6y^2}=\dfrac{3azx-6cyz}{9z^2}=0$
$\to \begin{cases}2bxy-3azx=0\\6cyz-2bxy=0\\3azx-6cyz=0\end{cases}$
$\to \begin{cases}2bxy=3azx\\6cyz=2bxy\\3azx=6cyz\end{cases}$
$\to 3azx=2bxy=6cyz$
$\to \dfrac{3azx}{6xyz}=\dfrac{2bxy}{6xyz}=\dfrac{6cyz}{6xyz}$
$\to \dfrac{a}{2y}=\dfrac{b}{3z}=\dfrac{c}{x}$
