Question

Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

CMR:\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)

Answer 1

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Answer 2

Là sao?

Answer 3

Từ\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}\Rightarrow4b^2z-6bcy=3acx-a^2z\)

\(\Rightarrow\)\(\left(4b^2+a^2\right)z=3c\left(ax+2by\right)\)

\(\Rightarrow\frac{z}{3c}=\frac{ax+2by}{a^2+4b^2}=\frac{ax}{a^2}=\frac{2by}{4b^2}\Rightarrow\frac{z}{3c}=\frac{x}{a}=\frac{y}{2b}\)

Answer 4

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