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\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{b}{y}=\frac{c}{z}\\\frac{c}{z}=\frac{a}{x}\\\frac{a}{x}=\frac{b}{y}\end{cases}}\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{z}{c}\)
\(\Leftrightarrow x:y:z=a:b:c\)
Ta có: \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
=> \(\frac{a\left(bz-cy\right)}{a^2}=\frac{b\left(cx-az\right)}{b^2}=\frac{c\left(ay-bx\right)}{c^2}\)
=> \(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{c^2+b^2+c^2}=0\)
=> \(\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}}\) => \(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}}\) => \(\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\) => \(\hept{\begin{cases}\frac{b}{y}=\frac{c}{z}\\\frac{c}{z}=\frac{a}{x}\\\frac{a}{x}=\frac{b}{y}\end{cases}}\) => \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)=> \(a:b:c=x:y:z\)
Ta có :
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}\)
\(=\frac{ayz-bxz}{cz}=\frac{0}{ax+by+cz}=0\)
\(\Leftrightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\) \(\left(1\right)\)
\(cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\) \(\left(2\right)\)
\(ay=bx\Rightarrow\frac{y}{b}=\frac{x}{a}\) \(\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) hay \(x:y:z=a:b:c\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abx-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
=>bz-cy=cx-az=ay-bx=0
- bz-cy=0 => bz=cy => \(\frac{b}{y}=\frac{c}{z}\)
- cx-az=0 => cx=az => \(\frac{c}{z}=\frac{a}{x}\)
=>\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow a:b:c=x:y:z\)(đpcm)
Ta có \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Rightarrow\frac{\left(bz-cy\right).x}{ax}=\frac{\left(cx-az\right)y}{by}=\frac{\left(ay-bx\right).z}{cz}\)
\(\Rightarrow\frac{bxz-cxy}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}\)
Suy ra:
bz - cy = 0 (1)
cx - az = 0 (2)
ay - bx = 0 (3)
Từ (1) ta có: \(bz=cy\Rightarrow\frac{y}{b}=\frac{z}{c}\left(I\right)\)
Từ (2) ta có: \(cx=az=\frac{z}{c}=\frac{x}{a}\left(II\right)\)
Từ (3) ta có: \(ay=bx=\frac{x}{a}=\frac{y}{b}\left(III\right)\)
Từ (I), (II), (III) => x: y: z = a: b: c
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
=>\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
=>\(\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}\)
hay x:y:z=a:b:c
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