Vì \(x=2018\Rightarrow x+1=2019\)
Thay x+1=2019 vào biểu thức A ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)
\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)
\(=1\)
\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)
\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)
\(-2018x-x+2019\)
\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)
\(+x\left(x-2018\right)-\left(x-2018\right)+1\)
= 1
Vì \(x=2018\Rightarrow x+1=2019\)
Thay \(x+1=2019\) vào biểu thức \(A\) ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x-1\right)\)
\(=x^6-x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x+1\)
\(=1\)
A = x6 - 2019.x5 + 2019.x4 - 2019.x3 + 2019x2 - 2019x + 2019
Có: f(x) = x6 - 2019x5 + 2019x4 - 2019x3 + 2019x2 - 2019x + 2019
f(x) = x6 - (2018 + 1)x5 + (2018 + 1)x4 - ... - (2018 + 1)x + 2019
f(x) = x6 - (x + 1)x5 + (x + 1)x4 - ... - (x + 1)x + 2019
f(x) = x6 - (x6 + x5) + (x5 + x4) - ... - (x2 + x) + 2019
f(x) = x6 - x6 + x5 + x5 + x4 - ... - x2 + x + 2019
f(x) = -x + 2019
Thay x = 2018 vào f(x), ta có:
f(2018) = -2018 + 2019
= 1
=> f(2018) = 1
