x²⁰¹⁹-2019.x²⁰¹⁸+2019.x²⁰¹⁸+2019.x²⁰¹⁷-2019.x²⁰¹⁶+......2019.x-200     Tại x=2018

Giúp mik vs nhé 

Sai đề nên t sửa luôn nhé!

Vì \(x=2018\Rightarrow2019=2018+1=x+1\)

\(A=x^{2017}-2019\cdot x^{2018}+2019\cdot x^{2017}-2019\cdot x^{2016}+....+2019\cdot x-200\)

\(\Rightarrow A=x^{2019}-\left(x+1\right)x^{2018}+\left(x+1\right)x^{2017}-\left(x+1\right)x^{2016}+....-\left(x+1\right)x^2+\left(x+1\right)x-200\)

\(\Rightarrow A=x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-x^{2017}-x^{2016}+....-x^3-x^2+x^2+x-200\)

\(\Rightarrow A=x-200=2018-200=1818\)

Đề bài yêu cầu gì?

B=\(x^{2019}-2019.x^{2018}+2019.x^{2017}-...+2019x-1\)

Ta có : 2019 = 1+2018=1+x ( vì x = 2018 )

Suy ra : \(x^{2019}-\left(x+1\right).x^{2018}+\left(x+1\right).x^{2017}-....+\left(x+1\right).x-1\)

=\(x^{2019}-\left(x^{2019}+x^{2018}\right)+\left(x^{2018}+x^{2017}\right)-...+\left(x^2+x\right)-1\)

= \(x^{2019}-x^{2019}-x^{2018}+x^{2018}+x^{2017}-....+x^2+x-1\)

= \(x-1\) mà x =2018

=> \(x-1=2018-1=2017\)

Vậy giá trị của biểu thức B = 2017