t thì chẳng thấy dễ chút nào nhưng t làm dc
Ta thấy a, b, c \(\ne\) 0 nên a2 + (2b)2 + (3c)2 \(\ne\) 0.
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}=\frac{2abz-3cay}{a^2}=\frac{6bcx-2abz}{\left(2b\right)^2}=\frac{3cay-6bcx}{\left(3c\right)^2}\)
\(=\frac{2abz-3cay+6bcx-2abz+3cay-6bcx}{a^2+\left(2b\right)^2+\left(3c\right)^2}\) (Do a2 + (2b)2 + (3c)2 \(\ne\) 0)
\(=\frac{0}{a^2+\left(2b\right)^2+\left(3c\right)^2}=0\)
\(\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{y}{2b}=\frac{z}{3c}\\\frac{z}{3c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{2b}\end{matrix}\right.\)
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
