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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:
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=\frac{\left(2abz-2abz\right)-\left(3acy-3acy\right)+\left(6bcx-6bcx\right)}{a^2+4b^2+9c^2}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{matrix}\right.\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\left(đpcm\right).\)
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: \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(\Rightarrow\frac{2bzx-3cyx}{ax}=\frac{3cxy-azy}{2by}=\frac{ayz-2bxz}{3xz}\)
\(=\frac{2bzx-3cyx-3cxy-azy-ayz-2bxz}{ax-2by-3xz}=0\)
\(\Rightarrow\)\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}=0\)
\(\Rightarrow2bz=3cy;\)\(3cx=az;\)\(ay=2bx\)
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\).
a, xy+2x-y=5
=> x(y+2)-y-2=3
=>x(y+2)-(y+2)=3
=>(x-1)(y+2)=3
=>\(\hept{\begin{cases}x-1=3\Rightarrow x=4\\y+2=1\Rightarrow y=-1\end{cases}}\); \(\hept{\begin{cases}x-1=1\Rightarrow x=2\\y+2=3\Rightarrow y=1\end{cases}}\)
=>\(\hept{\begin{cases}x-1=-1\Rightarrow x=0\\y+2=-3\Rightarrow y=-5\end{cases}}\); \(\hept{\begin{cases}x-1=-3\Rightarrow x=-2\\y+2=-1\Rightarrow y=-3\end{cases}}\)
vậy (x;y)\(\in\)(4,-1);(2,1);(0,-5);(-2.-3)
từ\(\frac{2bz-3cy}{a}\)=\(\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
=>\(\frac{2abz-3acy}{a}\)=\(\frac{6bcx-2abz}{2b}\)=\(\frac{3cay-6cbx}{3c}\)
=\(\frac{2abz-3acy+6bcx-2abz+3cay-6cbx}{2a+4b+6c}\)=0
=>\(\frac{2bz-3cy}{a}=0\)=>2bz=3cy=>\(\frac{z}{3c}\)=\(\frac{y}{2b}\)(1)
=>\(\frac{3cx-az}{2b}\)=0 =>3cx=az =>\(\frac{x}{a}\)=\(\frac{z}{3c}\)(2)
=>\(\frac{ay-2bx}{3c}=0\)=>ay=2bx =>\(\frac{y}{2b}\)=\(\frac{x}{a}\)(3)
Từ (1),(2) và (3) suy ra\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)đpcm