## 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

Question

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

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4 tuần 2021-07-11T00:59:02+00:00 2 Answers 3 views 0

1. (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

2. (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