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

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    0
    2021-07-11T01:00:20+00:00

    `(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`

     

    0
    2021-07-11T01:00:37+00:00

    `(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`

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