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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=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
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
\(f\left(x\right)=x^5-2009x^4+2009x^3-2009x^2+2009x-2010\)
\(f\left(2008\right)=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-2010\)
\(f\left(2008\right)=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-2010\)
\(f\left(2008\right)=x-2010=2008-2010=-2\)
\(a,\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)=\dfrac{2}{3}.\dfrac{2}{4}=\dfrac{1}{3}\)
\(b,2.\left(\dfrac{-3}{2}\right)-\dfrac{7}{2}=-6.\dfrac{1}{2}-7.\dfrac{1}{2}=\left(-6-7\right).\dfrac{1}{2}=-13.\dfrac{1}{2}=\dfrac{-13}{2}\)
\(c,-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}=-\dfrac{3}{4}.\left(\dfrac{68}{13}-\dfrac{36}{13}\right)=-\dfrac{3}{4}.\dfrac{32}{13}=-\dfrac{24}{13}\)
a) \(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}\)
\(=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{2}{3}.\dfrac{2}{4}\)
\(=\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{3}\)
b) \(2.\left(\dfrac{-3}{2}\right)^2-\dfrac{7}{2}\)
\(=2.\dfrac{9}{4}-\dfrac{7}{2}\)
\(=\dfrac{9}{2}-\dfrac{7}{2}\)
\(=\dfrac{2}{2}=1\)
c) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
\(=-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
\(=\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
\(=\dfrac{3}{4}.\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
\(=\dfrac{3}{4}.\dfrac{-104}{13}\)
\(=\dfrac{3}{4}.\left(-8\right)\)
\(=-6\)
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..............
\(A=x^{100}-2009x^{99}+2008x^{98}-\ldots-2009x+4017\)
\(=x^{100}-2008x^{99}-x^{99}+2008x^{98}+x^{98}-\ldots-x+4017\)
thay x= 2008
Ta thấy các số hạng triệt tiêu nhau
\(A=-2008+4017\)
\(A=2009\)
x=2008 nên x+1=2009
Sửa đề: \(A=x^{100}-2009x^{99}+2009x^{98}-...+2009x^2-2009x+4017\)
\(=x^{100}-x^{99}\left(x+1\right)+x^{98}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+4017\)
\(=x^{100}-x^{100}-x^{99}+x^{99}+...+x^3+x^2-x^2-x+4017\)
=-x+4017
=-2008+4017
=2009