Giải:
\(\dfrac{bz-xy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)
\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)
\(\Rightarrow\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}\)
\(=0\)
Ta có: \(\dfrac{bz-cy}{a}\)
\(\Rightarrow bz-cy=0\)
\(\Rightarrow\dfrac{z}{c}=\dfrac{b}{y}\)\(\left(1\right)\)
Ta có:\(\dfrac{cx-az}{b}=0\)
\(\Rightarrow cx-az=0\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{z}{c}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)(đpcm)
Mình đã làm ở đây: Câu hỏi của Huyền Trang Tiến Tài
