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`