Ta có:
\(\dfrac{14bz-15cy}{a}=\dfrac{9cx-7az}{b}=\dfrac{5ay-6bx}{c}\)
\(\Rightarrow\dfrac{14abz-15acy}{a^2}=\dfrac{18bcx-14abz}{2b^2}=\dfrac{15acy-18bcx}{3c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{14abz-15acy}{a^2}=\dfrac{18bcx-14abz}{2b^2}=\dfrac{15acy-18bcx}{3c^2}=\dfrac{14abz-15acy+18bcx-14abz+15acy-18bcx}{a^2+2b^2+3c^2}=0\)
\(\Rightarrow\left\{{}\begin{matrix}14bz=15cy\\9cx=7az\\5ay=6bx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5y}{2b}=\dfrac{7z}{3c}\\\dfrac{3x}{a}=\dfrac{7z}{3c}\\\dfrac{3x}{a}=\dfrac{5y}{2b}\end{matrix}\right.\Leftrightarrow\dfrac{3x}{a}=\dfrac{5y}{2b}=\dfrac{7z}{3c}\)
