Câu hỏi của Lê Trọng Chương

Question

Cho các số a,b,c khác 0 thỏa mãn $\frac{bf-ce}{a}=\frac{cd-af}{b}=\frac{ae-bd}{c}$. Chứng minh rằng:$\frac{a}{d}=\frac{b}{e}=\frac{c}{f}$...

Đinh quang hiệp · Accepted Answer

$\frac{bf-ce}{a}=\frac{cd-àf}{b}=\frac{ae-bd}{c}=\frac{abf-ace}{a^2}=\frac{bcd-abf}{b^2}=\frac{ace-bcd}{c^2}$

$=\frac{abf-ace+bcd-abf+ace-bcd}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$

$\Rightarrow\frac{bf-ce}{a}=\frac{cd-af}{b}=\frac{ae-bd}{c}=0$

$\Rightarrow bf-ce=0\Rightarrow bf=ce\Rightarrow\frac{b}{e}=\frac{c}{f}\left(1\right)$

$cd-af=0\Rightarrow cd=af\Rightarrow\frac{c}{f}=\frac{a}{d}\left(2\right)$

$ae-bd=0\Rightarrow ae=bd\Rightarrow\frac{a}{d}=\frac{b}{e}\left(3\right)$

từ $\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{a}{d}=\frac{b}{e}=\frac{c}{f}$

Câu trả lời:

$\frac{bf-ce}{a}=\frac{cd-àf}{b}=\frac{ae-bd}{c}=\frac{abf-ace}{a^2}=\frac{bcd-abf}{b^2}=\frac{ace-bcd}{c^2}$

$=\frac{abf-ace+bcd-abf+ace-bcd}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$

$\Rightarrow\frac{bf-ce}{a}=\frac{cd-af}{b}=\frac{ae-bd}{c}=0$

$\Rightarrow bf-ce=0\Rightarrow bf=ce\Rightarrow\frac{b}{e}=\frac{c}{f}\left(1\right)$

$cd-af=0\Rightarrow cd=af\Rightarrow\frac{c}{f}=\frac{a}{d}\left(2\right)$

$ae-bd=0\Rightarrow ae=bd\Rightarrow\frac{a}{d}=\frac{b}{e}\left(3\right)$

từ $\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{a}{d}=\frac{b}{e}=\frac{c}{f}$