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\(f\left(x\right)=x^5-2009x^4+2009x^3-2009x^2+2009x-2010\)
\(f\left(2008\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2010\)
\(f\left(2008\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2010\)
\(f\left(2008\right)=x-2010=2008-2010=-2\)
Thay x=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
đặt 2009=x+1 ta đc:
\(x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2010=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2010=x-2010=2008-2010=-2\)
vậy..............
x=2010 nên x-1=2009
\(M=x^{2010}-x^{2009}\left(x-1\right)-...-x^2\left(x-1\right)-x\left(x-1\right)-1\)
\(=x^{2010}-x^{2010}+x^{2009}-x^{2009}+...-x^3+x^2-x^2+x-1\)
=x-1
=2009
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!!!!
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
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
ok chưa vậy