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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:
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=\frac{\left(2abz-2abz\right)-\left(3acy-3acy\right)+\left(6bcx-6bcx\right)}{a^2+4b^2+9c^2}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{matrix}\right.\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\left(đpcm\right).\)
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\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
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: \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(\Rightarrow\frac{2bzx-3cyx}{ax}=\frac{3cxy-azy}{2by}=\frac{ayz-2bxz}{3xz}\)
\(=\frac{2bzx-3cyx-3cxy-azy-ayz-2bxz}{ax-2by-3xz}=0\)
\(\Rightarrow\)\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}=0\)
\(\Rightarrow2bz=3cy;\)\(3cx=az;\)\(ay=2bx\)
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\).
Ta có :
\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
Áp dụng t/c của DTSBN ta có :
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2baz}{4b^2}=\frac{3cay-6cbx}{9c^2}\)\(=\frac{2abz-3acy+6bcx-2baz+3cay-6cbx}{a^2+4b^2+9c^2}\) \(=\frac{0}{a^2+4b^2+9c^2}=0\)
Suy ra :
+) \(\frac{2bz-3cy}{a}=\frac{2abz-3acy}{a^2}=0\)\(\Rightarrow\)2bz = 3cy \(\Rightarrow\)\(\frac{z}{3c}=\frac{y}{2b}\) (1)
+) \(\frac{ay-2bx}{3c}=\frac{3cay-6cbx}{9c^2}=0\)\(\Rightarrow\)ay = 2bx \(\Rightarrow\)\(\frac{y}{2b}=\frac{x}{a}\) (2)
Từ (1) và (2) suy ra \(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
đc đấy