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giả sử
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
ta có:\(\text{}\text{}\text{}\text{}\text{}\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cyx}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bxz}{cz}=\frac{bxz-cyx+cxy-ayz+ayz-bxz}{ax+by+cz}=0\)
\(\frac{bz-cy}{a}=0\Rightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\left(1\right)\)
\(\frac{cx-az}{b}=0\Rightarrow cx=az\Rightarrow\frac{z}{c}=\frac{x}{a}\left(2\right)\)
\(\frac{ay-bx}{c}=0\Rightarrow ay=bx\Rightarrow\frac{x}{a}=\frac{y}{b}\left(3\right)\)
từ (1),(2),(3) => \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
=> điều giả sử đúng => đpcm
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)
Ta có: \(\frac{bz-cy}{a}=\frac{bck-bck}{a}=0\left(1\right)\)
\(\frac{cx-az}{y}=\frac{cak-cak}{y}=0\left(2\right)\)
\(\frac{ay-bx}{c}=\frac{abk-abk}{c}=0\left(3\right)\)
Từ (1),(2),(3) => đpcm
Ta có :
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bxz}{cz}=\frac{0}{ax+by+cz}=0\)
Suy ra :
\(bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\) (1)
\(cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\) (2)
\(ay=bx\Rightarrow\frac{y}{b}=\frac{x}{a}\) (3)
Từ (1), (2), (3) suy ra \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)hay x : y : z = a : b : c.
Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
\(\Rightarrow\frac{bz-cy}{a}=0\Rightarrow bz-cy=0\Rightarrow bz=cy\). Hay \(\frac{b}{y}=\frac{c}{z}\) \((1)\)
\(\Rightarrow\frac{cx-az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\). Hay \(\frac{c}{z}=\frac{a}{x}\)\((2)\)
...
\(\frac{x}{a}=\frac{y}{b}\Rightarrow bx=ay\Rightarrow ay-bx=0\Rightarrow\frac{ay-bx}{c}=0\left(1\right)\)
\(\frac{x}{a}=\frac{z}{c}\Rightarrow cx=az\Rightarrow cx-az=0\Rightarrow\frac{cx-az}{b}=0\left(2\right)\)
\(\frac{y}{b}=\frac{z}{c}\Rightarrow cy=bz\Rightarrow bz-cy=0\Rightarrow\frac{bz-cy}{a}=0\left(3\right)\)
\(\text{Từ (1);(2) và (3) suy ra: }\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=0\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow\frac{bz-cy}{a}=0\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\left(1\right)\)
\(\frac{cx+az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\) \(=\frac{x\left(bz-cy\right)}{ax}=\frac{y\left(cx-az\right)}{by}=\frac{z\left(ay-bx\right)}{cz}\)
\(=\frac{bzx-cyx}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}=\frac{bzx-cyx+cxy-azy+ayz-bxz}{ax+by+cz}=0\)
\(\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{b}{y}=\frac{c}{z}\) (1)
\(\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\frac{c}{z}=\frac{a}{x}\) (2)
\(\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow\frac{a}{x}=\frac{b}{y}\) (3)
Từ (1);(2);(3) \(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)