cho a,b,c thuộc R biết a^2 + b^2 + c^2 = ab + bc +ca .Tính A = (a-b)^2015 + (b-c)^2016 + (c-a)^2017

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cho a,b,c thuộc R biết a^2 + b^2 + c^2 = ab + bc +ca .Tính A = (a-b)^2015 + (b-c)^2016 + (c-a)^2017

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Adeline 3 tuần 2021-09-06T06:00:05+00:00 2 Answers 8 views 0

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    0
    2021-09-06T06:01:13+00:00

    Có:

    `a^2 + b^2 + c^2 = ab + bc +ca`

    `⇔2.(a^2 + b^2 + c^2) = 2.(ab + bc +ca)`

    `⇔2a^2 + 2b^2 + 2c^2 = 2ab + 2bc +2ca`

    `⇔2a^2 + 2b^2 + 2c^2 – (2ab + 2bc +2ca)=0`

    `⇔2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca=0`

    `⇔(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0`

    `⇔(a-b)^2+(b-c)^2+(c-a)^2=0`

    Có: `(a-b)^2 \ge 0 , (b-c)^2 \ge 0, (c-a)^2 \ge 0 `

    `⇒(a-b)^2+(b-c)^2+(c-a)^2 \ge 0`

    Dấu ”=” xảy ra khi:

    `(a-b)^2=0, (b-c)^2=0, (c-a)^2=0`

    `⇔a-b=0, b-c=0, c-a=0`

    `⇔a=b=c.`

    Thay `a=b=c` vào `A = (a-b)^{2015} + (b-c)^{2016} + (c-a)^{2017}`, ta có:

    `A = (a-a)^{2015} + (b-b)^{2016} + (c-c)^{2017}`

    `A=0.`

    Vậy với `a,b,c∈RR, a^2 + b^2 + c^2 = ab + bc +ca ⇒ A=0.`

    0
    2021-09-06T06:01:41+00:00

    Ta có: $a^2+b^2+c^2=ab+bc+ca$

    $⇒2a^2+2b^2+2c^2=2ab+2bc+2ca$

    $⇒(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$

    $⇔(a-b)^2+(b-c)^2+(c-a)^2=0$

    Mà $(a-b)^2;(b-c)^2;(c-a)^2≥0$

    $⇒(a-b)^2+(b-c)^2+(c-a)^2≥0$

    Mà $(a-b)^2+(b-c)^2+(c-a)^2=0$

    Nên dấu = xảy ra $⇔a-b=0;b-c=0;c-a=0$
    Khi đó $A=0^{2015}+0^{2016}+0^{2017}=0$

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