Chứng minh rằng nếu (a ²+b ²+c ²)(x ²+y ²+z ²) = (ax+by+cz) ² với x,y,z khác 0 thì a/x = b/y = c/z
Chứng minh rằng nếu (a ²+b ²+c ²)(x ²+y ²+z ²) = (ax+by+cz) ² với x,y,z khác 0 thì a/x = b/y = c/z
By Audrey
By Audrey
Chứng minh rằng nếu (a ²+b ²+c ²)(x ²+y ²+z ²) = (ax+by+cz) ² với x,y,z khác 0 thì a/x = b/y = c/z
`(a ²+b ²+c ²)(x ²+y ²+z ²) = (ax+by+cz) ² `
`⇔a^2x^2+a^2y^2+a^2z^2+b^2x^2+c^2x^2+y^2b^2+b^2z^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz`
`⇔a^2y^2+a^2z^2+b^2z^2+c^2x^2+c^2y^2+b^2x^2-2axby-2axcz-2bycz=0`
`⇔(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0`
điều hiển nhiên
`”=”`xẩy ra khi :
`a/x = b/y = c/z`
`(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2`
`=> a^2x^2 + a^2y^2 + a^2z^2 + b^2x^2 + b^2y^2 + b^2z^2 + c^2x^2 + c^2y^2 + c^2z^2 = a^2x^2 + b^2y^2 + c^2z^2 + 2axby + 2axcz + 2bycz`
`=> a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 = 2axby + 2axcz + 2bycz`
`=> a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 – 2axby – 2axcz – 2bycz = 0`
`=> (a^2y^2 – 2axby + b^2x^2) + (a^2z^2 – 2axcz + c^2x^2 ) + (b^2z^2 -2bycz + c^2y^2) = 0`
`=> (ay – bx)^2 + (az – cx)^2 + (bz – cy)^2 = 0`
`\forall a ; b ; c ; x ; y ; z` ta có :
`(ay-bx)^2 \ge 0`
`(az-cx)^2 \ge 0`
`(bz – cy)^2 \ge 0`
`=> (ay – bx)^2 + (az – cx)^2 + (bz – cy)^2 \ge 0`
Dấu `=` xảy ra
`<=>` $\left\{\begin{array}{l} (ay-bx)^2 = 0 \\ (az – cx)^2 = 0 \\ (bz – cy)^2 = 0 \end{array}\right.$
`<=>` $\left\{\begin{array}{l}
ay=bx \\
az=cx \\
bz = cy
\end{array}\right.$
`<=>` $\left\{\begin{array}{l}
a/x = b/y \\
a/x = c/z \\
b/y =c/z
\end{array}\right.$
`<=> a/x = b/y = c/z`
Vậy nếu `(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2` với `x,y,z \ne 0` thì `a/x = b/y = c/z`