Mình làm theo cách của mình học ở trường là như sau:
\(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3x}\)
= \(\dfrac{a.\left(2bz-3cy\right)}{a.a}=\dfrac{2b\left(3cx-az\right)}{2b.2b}=\dfrac{3c\left(ay-2bx\right)}{3x.3x}\)
=\(\dfrac{2abz-3acy}{a^2}=\dfrac{6cbx-2abz}{2b^2}=\dfrac{3cay-6cbx}{9c^2}\)
=\(\dfrac{2abz-3acy}{a^2}+\dfrac{6cbx-2abz}{2b^2}+\dfrac{3cay-6cbx}{9c^2}\)
=\(\dfrac{0}{a^2+4b^2+9c^2}=0\)
=> \(\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{z}{3c}=\dfrac{y}{2b}\\\dfrac{x}{a}=\dfrac{y}{2b}\end{matrix}\right.\)
=> \(\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}\)( ĐPCM)
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