không biết có đúng ko

ta có: 3000x⁹⁸ -3000x⁹⁸ +3000x⁹⁷ -3000x⁹⁷ +.....

=0+0+0+....

=>x⁹⁹ +3000x⁹⁸ -3000x⁹⁸ +3000x⁹⁷ -........+3000x+1

= x⁹⁹ +0+0+...+3000x+1

= x.x⁹⁸ +3000x+1

=x(x⁹⁸+3000)+1

thay x=299.Ta có

299(299⁹⁸+3000)+1

1000x+2000x+3000x=12000x

6000x=12000x

x=12000:6000

x=2

f(x)=x⁹⁹-3000.x⁹⁸+3000.x⁹⁷-...-3000x²+3000x-1

f(2009)=x⁹⁹-(x+1).x⁹⁸+(x+1).x⁹⁷-...-(x+1)x²+(x+1)x-1

=x⁹⁹-x⁹⁹-x⁹⁸+x⁹⁸+x⁹⁷-...-x³-x²+x²+x-1

=(x⁹⁹-x⁹⁹)+(-x⁹⁸+x⁹⁸)+(x⁹⁷-x⁹⁷)...+(-x²+x²)+x-1

=2009-1

=2008

2009+1=2010 mà bạn

đặt 3000=x+1 ta đc

F(x)=\(x^{98}-\left(x+1\right)x^{97}+\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1=x^{98}-x^{98}-x^{97}+x^{97}+x^{96}-x^{96}.....-x^3-x^2+x^2+x-1=x-1=2009-1=2008\)

vậy.......

Hai Yến Trương Thị: kc chi

Ta có : 2999=x => x⁹⁹-3000x⁹⁸⁺3000x⁹⁷-...+3000x-1

f(x) = x⁹⁹ - (x+1)x⁹⁸+(x+1)x⁹⁷-...+(x+1)x-1

=x⁹⁹-x⁹⁹-x⁹⁸+x⁹⁸+x⁹⁷-...x²+x-1=x-1=2999-1=2998

Vậy : f(2999)= 2998

tại sao là (x+1) vậy

Giải thích hộ mình nhé

\(=1+2+3+4+...+100\)

\(=\frac{100.101}{2}=5050\)

Ta thấy 

\(f\left(x\right):g\left(x\right)\)

\(\Rightarrow\left(x^{100}+x^{99}+x^{98}+x^5+2020\right):\left(x^2-1\right)\)

\(=\left(x^{98}+x^{97}+2x^{96}+2x^{95}+...2x^4+3x^3+2x^2+3x+2\right)\) có số dư là \(R\left(x\right)=3x+2022\)

\(\Rightarrow R\left(2021\right)=3.2021+2022=8085\)