Đáp án:
Giải thích các bước giải:
$\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}$
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$⇒\dfrac{4bzx-6cyx}{2ax}=\dfrac{6cxy-2azy}{4by}=\dfrac{2ayz-4bxz}{6cz}$
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$=\dfrac{4bzx-6cyx+6cxy-2azy+2ayz-4bxz}{2ax+4by+6zy}$
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$=\dfrac{0}{2ax+4by+6zy}=0$
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$⇒2bz=3cy ; 3cx=az ; ay=2bx$
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$⇒\dfrac{y}{2b}=\dfrac{z}{3c} ; \dfrac{x}{a}=\dfrac{z}{3x} ; \dfrac{x}{a}=\dfrac{y}{2b}$
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$⇒\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}$ (đpcm)
