Question

Cho các số a,b,c,x,y,z thỏa mãn điều kiện

\(\dfrac{x}{\text{a}}=\dfrac{y}{b}=\dfrac{z}{c}\)

CMR\(\dfrac{bz-cy}{\text{a}}=\dfrac{cx-\text{az}\text{ }}{b}=\dfrac{\text{a}y-bx}{c}\)

Answer 1

ta có: x/a = y/b =z/c =xa/a^2 =yb/b^2 =zc/c^2 = (ax+by+cz)/(a^2+b^2+c^2)
=>x/a = (ax+by+cz)/(a^2+b^2+c^2) (1)
mặt khác ta có: x/a=y/b=z/c <=> x^2/a^2 =y^2/b^2 =z^2/c^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2)
=>x^2/a^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2) (2)
từ (1) và (2) ta => (ax+by+cz)^2/(a^2+b^2+c^2)^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2)
=> (x^2+y^2+z^2).(a^2+b^2+c^2)=(ax+by+cz)^2 => đpcm

Answer 2

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)

\(\dfrac{bz-cy}{a}=\dfrac{b.ck-c.bk}{a}=\dfrac{0}{a}=0\)(1)

\(\dfrac{cx-az}{b}=\dfrac{c.ak-a.ck}{b}=\dfrac{0}{b}=0\)(2)

\(\dfrac{ay-bz}{c}=\dfrac{a.bk-b.ak}{c}=\dfrac{0}{c}=0\)(3)

từ (1),(2) và(3) suy ra \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\left(đpcm\right)\)