Đáp án+Giải thích các bước giải:
`a)A=x^6-2021x^5+2021x^4-2021x^3+2021x^2-2021x+2021`
`=(x^6-2020x^5)-(x^5-2020x^4)+(x^4-2020x^3)-(x^3-2020x^2)+(x^2-2020x)-(x-2020)+1`
`=[x^5(x-2020)-x^4(x-2020)+x^3(x-2020)-x^2(x-2020)+x(x-2020)-(x-2020)]+1`
`=(x-2020)(x^5-x^4+x^3-x^2+x-1)+1`
Thay `x=2020` vào `A` ta có:
`A=(2020-2020)(2020^5-2020^4+2020^3-2020^2+2020-1)+1`
`=0.(2020^5-2020^4+2020^3-2020^2+2020-1)+1`
`=0+1`
`=1`
Vậy `A=1` khi `x=2020`
`b)B=x^{10}+20x^{9}+20x{8}+....+20x^2+20x+20`
`=(x^{10}+19x^9)+(x^9+19x^8)+......+(x^3+19x^2)+(x^2+19x)+(x+19)+1`
`=[x^9(x+19)+x^8(x+19)+.....+x^2(x+19)+x(x+19)+(x+19)]+1`
`=(x+19)(x^9+x^8+....+x^2+x+1)+1`
Thay `x=-19` vào `B` ta có:
`B=(-19+19)[(-19)^9+(-19)^8+....+(-19)^2+(-19)+1)+1`
`=0.[(-19)^9+(-19)^8+....+(-19)^2+(-19)+1)+1`
`=0+1`
`=1`
Vậy `B=1` khi `x=-19`
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