Cho đa thức: f(x)=x^6 – 2011x^5 + 2011x^4 -2011x^3 + 2011x^2 – 2011x + 2016 f(2010)= bao nhiêu

Question

Cho đa thức:
f(x)=x^6 – 2011x^5 + 2011x^4 -2011x^3 + 2011x^2 – 2011x + 2016
f(2010)= bao nhiêu

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Mackenzie 3 tuần 2021-07-08T17:36:21+00:00 1 Answers 9 views 0

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    0
    2021-07-08T17:37:56+00:00

    Đáp án:

    Có : `f (2010)`

    `-> x = 2010`

    `-> x + 1=  2011` `(1)`

    Có : `f (x) = x^6 – 2011x^5 + 2011 x^4 – 2011x^3 + 2011x^2 – 2011x + 2016`

    Thay `(1)` vào ta được :

    `⇔ f (x) = x^6 – (x + 1) x^5 + (x + 1)x^4 – (x+1)x^3 + (x+1)x^2-(x+1)x +  2016`

    `⇔ f (x) = x^6 – x^6 – x^5 + x^5 + x^4 – x^4 – x^3 + x^3 + x^2 – x^2 – x + 2016`

    `⇔ f (x) = (x^6 – x^6) + (-x^5 + x^5) + (x^4 – x^4) + (-x^3 + x^3) + (x^2 – x^2) + (-x + 2016)`

    `⇔ f (x)  = -x + 2016`

    `⇔ f (2010) = -2010 + 2016`

    `⇔ f (2010) =6`

    Vậy `f (2010) = 6`

     

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