Question

\(\dfrac{bz-cy}{z}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\) (1) CMR: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) (*)

Answer 1

Lời giải:
Sửa đề: $z$ đầu tiên ở mẫu đổi thành $a$.

Ta có:

$\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}$

$=\frac{abz-cya}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}$

$=\frac{abz-cya+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$

$\Rightarrow bz-cy=cx-az=ay-bx=0$

$\Rightarrow bz=cy; cx=az; ay=bx$

$\Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

Ta có đpcm.