Question

Cho dãy tỉ số bằng nhau:\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

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

Answer 1

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

\(\Rightarrow\frac{2bza-3acy}{a^2}=\frac{6cxb-2bza}{4b^2}=\frac{3ayc-6bxc}{9c^2}\)

\(=\frac{2bza-3cya+6xbc-2bza+3ayc-6bxc}{a^2+4b^2+9c^2}\)

\(=0\)

\(\Rightarrow\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}=0\)

\(\Rightarrow2bz=3cy;3cx=az;ay=2bx\)

\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\left(đpcm\right)\)

Answer 2

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

\(\Rightarrow\frac{2bzx-3cyx}{ax}=\frac{3cxy-azy}{2by}=\frac{ayz-2bxz}{3xz}\)

\(=\frac{2bzx-3cyx-3cxy-azy-ayz-2bxz}{ax-2by-3xz}=0\)

\(\Rightarrow\)\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}=0\)

\(\Rightarrow2bz=3cy;\)\(3cx=az;\)\(ay=2bx\)

\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\).