\(P\left(x\right)=x^{2017}-2018x^{2017}+2018x^{2016}-...-2018x+1\)

Vì \(x=2017\)

\(\Leftrightarrow x+1=2018\)

Thay vào P(x) ta được :

\(P\left(x\right)=x^{2017}-x^{2017}\left(x+1\right)+x^{2016}\left(x+1\right)-...-x\left(x+1\right)+1\)

\(P\left(x\right)=x^{2017}-x^{2018}-x^{2017}+x^{2017}+x^{2016}-...-x^2-x+1\)

\(P\left(x\right)=-x^{2018}+1\)

\(P\left(x\right)=-2017^{2018}+1\)

F(x)=\(x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x+1.\)

x=2017=>2018=x+1 thay vào F(x) ta có:

F(x)=x+1=2018

pkm;lkml

Ta có: x=2017

nên x+1=2018

Ta có: \(P=x^{15}-2018x^{14}+2018x^{13}-2018x^{12}+...+2018x^3-2018x^2+2018x-2018\)

\(=x^{15}-\left(x+1\right)\cdot x^{14}+\left(x+1\right)\cdot x^{13}-\left(x+1\right)\cdot x^{12}+...+\left(x+1\right)\cdot x^3-\left(x+1\right)\cdot x^2+\left(x+1\right)\cdot x-\left(x+1\right)\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+...+x^3-x^3+x^2-x^2+x-x-1\)

=-1

@ 肖战Daytoy_1005 giup

Vì \(x=2017\Rightarrow x+1=2018\)

Thay \(x+1=2018\)vào biểu thức A ta được :

\(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+\left(x+1\right)\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)

\(=1\)

Tại x=2017 thì 2018 = x + 1 

Khí đó \(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+x+1\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)

\(=1\)

\(A=x^9-2018x^8+2018x^7-2018x^6+2016x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)

\(A=x^9-\left(2017+1\right)x^8+\left(2017+1\right)x^7-...+\left(2017+1\right)x-\left(2017+1\right)\)

\(A=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-...+\left(x+1\right)x-x-1\)

\(A=x^9-x^9-x^8+x^8+x^7-...+x^2+x-x-1\)

\(A=-1\)

Mk sửa lại đề. bn tham khảo nha!!!              

    \(x=2017\)\(\Rightarrow\)\(x+1=2018\)

Ta có:    \(A=x^9-2018x^8+2018x^7-2018x^6+2018x^5-2018x^4+2018x^3-2018x^2+2018x-2018\)

\(=x^9-\left(x+1\right)x^8+\left(x+1\right)x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\)

\(=x^9-x^9-x^8+x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x-1\)

\(=1\)

Lời giải:

Ta có:

\(A=x^5-2018x^4+2018x^3-2018x^2+2018x-1000\)

\(A=(x^5-2017x^4)-(x^4-2017x^3)+(x^3-2017x^2)-(x^2-2017x)+x-1000\)

\(A=x^4(x-2017)-x^3(x-2017)+x^2(x-2017)-x(x-2017)+x-1000\)

Tại \(x=2017\Rightarrow A=2017^4.0-2017^3.0+2017^2.0-2017.0+2017-1000\)

\(A=2017-1000=1017\)

tối nay mk phải nộp rùi

giúp mk vs!!