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

\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}.\)

\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=\frac{\left(2abz-2abz\right)-\left(3acy-3acy\right)+\left(6bcx-6bcx\right)}{a^2+4b^2+9c^2}=0.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)

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

Chúc bạn học tốt!

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

Suy ra: \(\frac{a.\left(2bz-3cy\right)}{a.a}=\frac{2b\left(3cx-az\right)}{2b.2b}=\frac{3c.\left(ay-2bx\right)}{3c.3c}\)

\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{3bcx-abz}{2b^2}=\frac{acy-2cbx}{3c^2}\)

Theo tính chất dãy tỉ số bằng nhau

\(\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+2b^2+3c^2}=\frac{0}{a^2+2b^2+3c^2}=0\)

\(\Rightarrow\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}}\)

=> đpcm

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

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

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

\(=0\)

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

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

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

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

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

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

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

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

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

https://goo.gl/xr4NMs

Là sao?

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

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

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