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f(2005)= 2005-2006.2005+2006.2005-2006.2005+2006.2005-2006.2005+2009
f(2005)=4 nha!
f(x) = x6 - 2009x5 + 2009x4 - 2009x3 + 2009x2 - 2009x + 2011
x = 2008 => 2009 = x + 1
=> f(2008) = f(x+1) = x6 - (x+1)x5 + (x+1)x4 - (x+1)x3 + (x+1)x2 - (x+1)x + 2011
= x6 - x6 - x5 + x5 + x4 - x4 - x3 + x3 + x2 - x2 - x + 2011
= -x + 2011 = -2008 + 2011 = 3
Vậy ...
\(\left(2x-8x+2\right)-\left(-4x+3x+8\right)\)
\(=\left(-6x+2\right)-\left(-x+8\right)\)
\(=-6x+2+x-8\)
\(=\left(-6x+x\right)+\left(2+8\right)\)
\(=\left(-5x\right)+10\)
\(\left(2x-8x+2\right)-\left(-4x+3x+8\right)\)
\(=2x-8x+2+4x-3x-8\)
\(=-5x-6\)
\(P\left(x\right)-Q\left(x\right)=2x-6x-2\)
\(\rightarrow(2x-5x-3)-Q\left(x\right)=-4x-2\)
\(\rightarrow\left(-3x-3\right)-Q\left(x\right)=-4x-2\)
\(\rightarrow Q\left(x\right)=\left(-3x-3\right)-\left(-4x-2\right)\)
\(\rightarrow Q\left(x\right)=-3x-3+4x+2\)
\(\rightarrow Q\left(x\right)=\left(4x-3x\right)+\left(2-3\right)\)
\(\rightarrow Q\left(x\right)=x-1\)
A = (- 1,25)3. (\(\dfrac{2}{5}\))3
A = (-1,25 . \(\dfrac{2}{5}\))3
A = (- \(\dfrac{2}{4}\))3
A = (-\(\dfrac{1}{2}\))3
A = \(\dfrac{-1}{8}\)