\(f\left(x\right)=x^{99}-3000x^{98}+3000x^{97}-...+3000x^3-3000x^2+3000-x\)
Tính \(f\left(2009\right)\)
cho đa thức:
\(f\left(x\right)=x^{99}-3000x^{98}+3000x^{97}-3000x^{96}+...-3000x^2+3000x-1\)
Tính f(2009)?
Tìm F(299) biết F(x)= x99+3000x98-3000x98+3000x97-...+3000x +1
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
Tìm x :
1000x + 2000x + 88y+ 3000x= 12000x + 88y
1000x+2000x+3000x=12000x
6000x=12000x
x=12000:6000
x=2
Tính \(T=\left(\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)X\left(\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}\right)-\left(\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\right)X\left(\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}\right)\)
cho đa thức f(x) =\(x^{99}-3000.x^{98}+3000.x^{97}-3000.x^{96}+...-3000.x^2+3000.x-1\)
tính f(2009)
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
Cho \(f\left(x\right)=\frac{100^x}{100^x+10}\), tính tổng:
\(S=f\left(\frac{1}{2009}\right)+f\left(\frac{2}{2009}\right)+f\left(\frac{3}{2009}\right)+...+f\left(\frac{2008}{2009}\right)\)
\(f\left(x\right)+f\left(1-x\right)=\frac{100^x}{100^x+100}+\frac{100^{1-x}}{100^{1-x}+100}\)
Nhân cả tử và mẫu của \(\frac{100^{1-x}}{100^{1-x}+100}\) với \(100^x\) ta được:
\(f\left(x\right)+f\left(1-x\right)=\frac{100^x}{100^x+100}+\frac{100}{100+100^x}=\frac{100^x+100}{100^x+100}=1\)
Vậy: \(S=f\left(\frac{1}{2009}\right)+f\left(\frac{2008}{2009}\right)+f\left(\frac{2}{2009}\right)+f\left(\frac{2007}{2009}\right)+...+f\left(\frac{1004}{2009}\right)+f\left(\frac{1005}{2009}\right)\)
\(S=1+1+1+...+1\) (có \(\frac{2008-1+1}{2}=1004\) số 1)
\(S=1004\)
Cho đa thức :
F(x)= x^99 - 3000.x^97+3000.x^96+......-3000.x^2+3000.x-1. Tính f(2009)?
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)=x^99-3000.x^98+3000.x^97-.....-3000.x^2+3000.x-1
tự làm là hạnh phúc của mỗi công dân.
Thiếu đề : Tính f(2009)
\(F\left(x\right)=x^{98}-\left(x+1\right)x^{97}+\left(x+1\right)x^{96}+...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{98}-x^{98}-x^{97}+x^{97}+x^{96}-x^{96}...-x^3-x^2+x^2+x-1\)
\(=x-1=2009-1=2008\)
Tìm dư của phép chia:
a) \(f\left(x\right)=x^{100}+x^{99}+x^{98}+...+x+1\)chia cho \(g\left(x\right)=x-1\)
b) \(f\left(x\right)=100x^{100}-99x^{99}+98x^{98}+...+2x^2-x+1\)chia cho \(g\left(x\right)=x+1\)
a)\(f\left(x\right)=x^{100}+x^{99}+x^{98}+...+x+1\)chia cho \(g\left(x\right)=x-1\)
Ta có:\(f\left(x\right)=x^{100}+x^{99}+x^{98}+...+x+1\)
\(=x^{99}\left(x-1\right)+x^{98}\left(x-1\right)+...+\left(x-1\right)-99x+2\)
Vì x-1 chia hết cho x-1 nên \(x^{99}\left(x-1\right)+x^{98}\left(x-1\right)+...+\left(x-1\right)\)chia hết cho x-1
Do đó \(x^{99}\left(x-1\right)+x^{98}\left(x-1\right)+...+\left(x-1\right)-99x+2\) cha x-1 dư 2-99x
Vậy \(f\left(x\right)=x^{100}+x^{99}+x^{98}+...+x+1\)chia cho \(g\left(x\right)=x-1\) dư 2-99x
Không biết có đúng ko nữa
a/ Trước tiên ta chứng minh với mọi số tự nhiên \(n\ge1\)
\(x^n-1⋮\left(x-1\right)\)điều này dễ chứng minh nên mình bỏ qua nhé.
Ta có:
\(f\left(x\right)=x^{100}+x^{99}+...+x+1\)
\(=\left(x^{100}-1\right)+\left(x^{99}-1\right)+...+\left(x-1\right)+101\)
Vậy f(x) chia cho g(x) dư 101.
b/ \(f\left(x\right)=100x^{100}-99x^{99}+...-x+1\)
\(=100\left(x^{100}+x^{99}\right)-\left(100+99\right)\left(x^{99}+x^{98}\right)+...-\left(100+99+...+1\right)\left(x+1\right)+\left(100+99+...+1\right)+1\)
\(=100x^{99}\left(x+1\right)-\left(100+99\right)x^{98}\left(x+1\right)+...-\left(100+99+...+1\right)\left(x+1\right)+\left(100+99+...+1\right)+1\)
Từ đây ta có số dư của f(x) cho g(x) là:
\(\left(100+99+...+1\right)+1=\dfrac{100.101}{2}+1=5051\)