f(x) = \(\left(x^6-2019x^5\right)-\left(x^5-2019x^4\right)+\left(x^4-2019x^3\right)-\left(x^3-2019x^2\right)+\left(x^2-2019x\right)-\left(x-2019\right)+1\)

= \(x^5\left(x-2019\right)-x^4\left(x-2019\right)+x^3\left(x-2019\right)-x^2\left(x-2019\right)+x\left(x-2019\right)-\left(x-2019\right)+1\)

Thay x = 2019 vào f(x), ta có:

f(2019) = 0 + 0 + 0 + 0 + 0 +0 + 1 = 1

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

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

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

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

\(=-19\)

\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)

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

\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)

\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)

\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)

\(=-1\)

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 \(f\left(2019\right)=2018\)

a,x²-8x=0

⇔x(x-8)=0

⇔x=0 hoặc x-8=0

⇔x=0 hoặc x=8

Vậy tập nghiệm của phương trình đã cho là:S={0;8}

b,x²-2020x+2019=0

⇔x²-2019x-x+2019=0

⇔x(x-2019)-(x-2019)=0

⇔(x-2019)(x-1)=0

⇔x-2019=0 hoặc x-1=0

⇔x=2019 hoặc x=1

Vậy tập nghiệm của phương trình đã cho là:S={2019;1}

c,(2x-1)²-(x+5)²=0

⇔(2x-1-x-5)(2x-1+x+5)=0

⇔(x-6)(3x+4)=0

⇔x-6=0 hoặc 3x+4=0

⇔x=6 hoặc x=\(\frac{-4}{3}\)

Vậy tập nghiệm của phương trình đã cho là:S={6;\(\frac{-4}{3}\)}

Chứng minh hay giải PT??