Theo đầu bài ta có :\(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\)
Lại có a,b,c\(\ne\)0 vì mẫu phải khác 0
=>\(\dfrac{2bz-3cy}{a}.\dfrac{a}{a}=\dfrac{3cx-az}{2b}.\dfrac{2b}{2b}=\dfrac{ay-2bx}{3c}.\dfrac{3c}{3c}\)
=>\(\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có :
\(\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}=\dfrac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=\dfrac{0}{a^2+4b^2+9c^2}=0\)
\(\dfrac{2abz-3acy}{a^2}=0\Rightarrow2abz=3acy\) => 2bz = 3cy => \(\dfrac{z}{3c}=\dfrac{y}{2b}\) (1)
\(\dfrac{6bcx-2abz}{4b^2}=0\) => 6bcx = 2abz => 3cx = az => \(\dfrac{x}{a}=\dfrac{z}{3c}\) (2)
Từ (1) và (2) =>\(\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}\) (đpcm)
