Đáp án+Giải thích các bước giải:
`f(x)=x^6-2020x^5+2020x^4-2020x^3+2020x^2-2020x+2020`
`=x^6-(2019+1).x^5+(2019+1).x^4-(2019+1).x^3+(2019+1).x^2-(2019+1).x+2019+1`
`=x^6-2019x^5-x^5+2019x^4+x^4-2019x^3-x^3+2019x^2+x^2-2019x-x+2019+1`
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Ta có: `f(2019)=2019^6-2019.2019^5-2019^5+2019.2019^4+2019^4-2019.2019^3-2019^3+2019.2019^2+2019^2-2019.2019-2019+2019+1`
`=(2019^6-2019^6)+(-2019^5+2019^5)+(2019^4-2019^4)+(-2019^3+2019^3)+(2019^2-2019^2)+(-2019+2019)+1`
`=1`
Vậy `f(2019)=1`
