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Thay x=2008 vao cac thua so 2009 trong da thuc duoc :
x9 - (x+1)x8 +(x+1)x7 - (x+1)x6 + (x+1)x5 - (x+1)x4 + (x+1)x3 - (x+1)x2 + (x+1)x +(x+1)
=x9 - x9 - x8 + x8 + x7 - x7 - x6 + x6 + x5 - x5 - x4 + x4 + x3 - x3 - x2 + x2 + x + x +1
= 2x + 1= 4017
Ta có: x=1999
nên x+1=2020
Ta có: \(f\left(x\right)=x^{17}-2020\cdot x^{16}+2020\cdot x^{15}-2020\cdot x^{14}+...+2000x-1\)
\(=x^{17}-x^{16}\left(x+1\right)+x^{15}\left(x+1\right)-x^{14}\left(x+1\right)+...+x\left(x+1\right)-1\)
\(=x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-x^{15}-x^{14}+...+x^2+x-1\)
\(=x-1\)
\(=1999-1=1998\)
f(x) = x^17 - 2000x^16 + 2000x^15 - 2000x^14 + ... + 2000x - 1
⇒ f(1999) = 1999^17 - 2000.1999^16 + 2000.1999^15 - 2000.1999^14 + ... + 2000.1999 - 1
⇒ 1999. f(1999) = 1999^18 - 1999.1999^17 + 2000.1999^16 - 2000.1999^15 + ... + 2000.1999^2 - 1999
⇒ 1999. f(1999) + f(1999) =(1999^18 - 1999.1999^17 + 2000.1999^16 - 2000.1999^15 + ... + 2000.1999^2 - 1999) + (1999^17 - 2000.1999^16 + 2000.1999^15 - 2000.1999^14 + ... + 2000.1999 - 1)
⇒ 2000. f(1999) = 19992−1
⇔ f(1999) =1999^2-1/2000(ghi dưới dạng phân số nha)
x^5 - 2009x^4 + 2009x^3 - 2009x^2 + 2009x - 2010
= 2008^5 - 2009.2008^4 + 2009.2008^3 - 2009.2008^2 +2009.2008x - 2010
= 2008^5 - 2008.2008^4 - 1.2008^4 + 2008.2008^3 + 1.2008^3 - 2008.2008^2 - 1.2008^2 + 2008.2008 + 1.2008 -2010
= 2008^5 - 2008^5 -2008^4 + 2008^4 + 2008^3 - 2008^3 - 2008^2 + 2008^2 + 2008 - 2010
= 0 - 0 + 0 - 0 + ( - 2 )
=- 2
x=2008
nên x+1=2009
\(P\left(x\right)=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+...-x^3-x^2+x^2+x\)
=x=2008
Bài 1: 2008^5 - 2009.2008^4+2009.2008^3 - 2009.2008^2+2009.2008-2010
= 2008^5-(2008.2008^4-1.2008^4)+(2008.2008^3+1.2008^3)+(2008.2008^2-1.2008^2)+(2008.2008-1.2008)-2010
= 2008^5-(2008^5-2008^4)+(2008^4+2008^3)+(2008^3-2008^2)+ (2008^2+2008)-2010
= (2008^5-2008^5) + (-2008^4+2008^4)+ (2008^3-2008^3)+(-2008^2-2008^2)+(2008-2010)
=0+0+0+0+(-2)
=2
Tick mik nha!!!!
\(f\left(x\right)=x^5-2009x^4+2009x^3-2009x^2+2009x-2010\)
\(f\left(2008\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2010\)
\(f\left(2008\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2010\)
\(f\left(2008\right)=x-2010=2008-2010=-2\)
thay x = -1 vào biểu thức \(x^{2008}-x^{2007}+1\)
có: \(\left(-1\right)^{2008}-\left(-1\right)^{2007}+1\)
\(=1-\left(-1\right)+1\)
\(=3\)
CHÚC BN HỌC TỐT!!!!
