\(f\left(2019\right)=x^{100}-\left(2019+1\right)x^{99}+\left(2019+1\right)x^{98}-....+\left(2019+1\right)x^2-\left(2019+1\right)x+2000\)

\(=x^{100}-\left(x+1\right)x^{99}+\left(x+1\right)x^{98}-...+\left(x+1\right)x^2-\left(x+1\right)x+2000\)

\(=x^{100}-x^{100}-x^{99}+x^{99}+x^{98}-...+x^3+x^2-x^2-x+2000\)

\(=-x+2000=-2019+2000\)

\(=-19\)

cảm ơn ạ 😊😙

Bài làm:

Ta có: \(x=2019\Rightarrow2020=x+1\)

Thay vào ta được:

\(f\left(2019\right)=x^{99}-\left(x+1\right)x^{98}+\left(x+1\right)x^{97}-\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(f\left(2019\right)=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}+...-x^3-x^2+x^2+x-1\)

\(f\left(2019\right)=x-1\)

Thay \(x=2019\)vào ta được:

\(f\left(2019\right)=2019-1=2018\)

Vậy f(2019) = 2018

\(f\left(x\right)=x^{99}-2020x^{98}+2020x^{97}-2020x^{96}+...-2020x^2+2020x-1\)

\(f\left(2019\right)=2019^{99}-2020.2019^{98}+2020.2019^{97}-...+2020.2019-1\)

Xét  \(2020.2019^{98}=2019^{99}+2019^{98};2020.2019^{97}=2019^{98}+2019^{97}\)

\(2020.2019^{96}=2019^{97}+2019^{96};...;2020.2019=2019^2+2019\)

\(\Rightarrow f\left(2019\right)=2019^{99}-2019^{99}-2019^{98}+2019^{97}-2019^{97}-...+2019^2+2019-1\)

\(\Rightarrow f\left(2019\right)=2019-1=2018\). Vậy \(f\left(2019\right)=2018\)

b) Ta có : \(x=2019\) \(\Rightarrow x+1=2020\) Thay vào biểu thức ta được :

( Chỗ nào có 2020 thay thành x + 1 )

\(x^9-\left(x+1\right).x^8+\left(x+1\right).x^7-....-\left(x+1\right).x^2+\left(x+1\right).x\)

\(=x^9-x^9-x^8+x^8+x^7-...-x^3-x^2+x^2+x\)

\(=x\\ \)

\(=2019\)

Vậy : biểu thức trên bằng 2019 với x = 2019.

Dễ thấy \(VT\ge0\Rightarrow2020x\ge0\Leftrightarrow x\ge0\)

\(\Rightarrow pt\Leftrightarrow2019x+\frac{2019.2020}{2}=2020x\)

\(\Leftrightarrow x=2019.1010\)