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.......

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

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

ta có:3000=2009+1 \Rightarrow f(2009)=200999200999-3000.200998200998+3000.200997200997-3000.200996200996+...+3000.2009-1 =200999200999-(2009+1)200998200998+(2009+1).200997200997-(2009+1).200996200996+...+(2009+1).2009-1 =200999200999-(200999200999+200998200998)+(200998200998+200997200997)-(200997200997)+.200996200996)+...+(2009220092+2009)-1 =200999200999-200999200999-200998200998+200998200998+200997200997-200997200997-200996200996+...+2009220092+2009-1 =2009-1 =2008

tự làm là hạnh phúc của mỗi công dân.

Thiếu đề : Tính f(2009)

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

\(=x^{98}-x^{98}-x^{97}+x^{97}+x^{96}-x^{96}...-x^3-x^2+x^2+x-1\)

\(=x-1=2009-1=2008\)

3000 :2=?

Xét mẫu :

\(\frac{2999}{1}+\frac{2998}{2}+.....+\frac{1}{2999}\)

=\(\left(1+\frac{2998}{2}\right)+\left(1+\frac{2997}{3}\right)+....+\left(1+\frac{1}{2999}\right)+1\)

=\(\frac{3000}{2}+\frac{3000}{3}+.....+\frac{3000}{2999}+\frac{3000}{3000}\)

=\(3000\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{3000}\right)\)

Thay vào ta có:

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{3000}\right)}\)

=\(\frac{1}{3000}\)

Học sinh nhẩm và ghi kết quả như sau:

3000 + 2000 x 2 = 7000

(3000 + 2000 ) x 2 = 10000