\(\dfrac{xn-ym}{p^2}=\dfrac{yp-zn}{m^2}=\dfrac{mz-xp}{n^2}\)
\(\Rightarrow\dfrac{xnp-ymp}{p^3}=\dfrac{ymp-znm}{m^3}=\dfrac{znm-xnp}{n^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{xnp-ymp}{p^3}=\dfrac{ymp-znm}{m^3}=\dfrac{znm-xnp}{n^3}=\dfrac{xnp-ymp+ymp-znm+znm-xnp}{p^3+m^3+z^3}=\dfrac{0}{p^3+m^3+z^3}=0\)
Nên \(\left\{{}\begin{matrix}xn=ym\\yp=zn\\mz=xp\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{m}=\dfrac{y}{n}\\\dfrac{y}{n}=\dfrac{z}{p}\\\dfrac{x}{m}=\dfrac{z}{p}\end{matrix}\right.\Leftrightarrow\dfrac{x}{m}=\dfrac{y}{n}=\dfrac{z}{p}\)
Hay \(x;y;z\) tỉ lệ với \(m;n;p\left(đpcm\right)\)