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Với x = 2005 ta có
\(x^{2005}-2006x^{2004}+2006x^{2003}-2006x^{2002}+...-2006x^2+2006x-1\)
\(=\left(x^{2005}-2005x^{2004}\right)-\left(x^{2004}-2005^{2003}\right)+\left(x^{2003}-2005x^{2002}\right)-...-\left(x^2-2005x\right)+\left(x-2005\right)+2006\)
\(=\left(x-2005\right)\left(x^{2004}-x^{2003}+x^{2002}-...-x+1\right)+2006=2006\).
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Ta có :
\(x^{2005}-2006x^{2004}+2006x^{2003}-...-2006x^2+2006x-1\)
\(=\left(x^{2005}-2005x^{2004}\right)-\left(x^{2004}-2005x^{2003}\right)+...-\left(x^2-2005x\right)+x-1\)
\(=\left(x-2005\right)\left(x^{2004}-x^{2003}-...-x\right)+x-1\)
\(=\left(2005-2005\right)\left(x^{2004}-x^{2003}-...-x\right)+2005-1\)
\(=2004\)
Ta có: x2005 − 2006x2004 + 2006x22003 − 2006x2002 + ...− 2006x2 + 2006x − 1
= x2005 − (x + 1)x2004 + (x + 1)x2003 + (x + 1)x2002 + ... - (x + 1)x2 + (x + 1)x - 1
= x2005 − x2005 - x2004 +x2004 + x2003 - x2003 - x2002 +.... - x3 - x2 + x2 +x - 1
= x - 1 = 2004
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a,\(A=x^{2005}-2006x^{2004}+............+2006x-1\\ A=x^{2005}-\left(x+1\right)x^{2004}+..............+\left(x+1\right)x-1\\ A=x^{2005}-x^{2005}+x^{2004}-x^{2004}+.............+x^2+x-1\\ A=x-1\\ \Leftrightarrow A=2004\)vậy
a,A=x2005−2006x2004+............+2006x−1A=x2005−(x+1)x2004+..............+(x+1)x−1A=x2005−x2005+x2004−x2004+.............+x2+x−1A=x−1⇔A=2004
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^{2005}-2005x^{2004}-x^{2004}+2005x^{2003}+x^{2003}-2005x^{2002}-.....+x^3-2005x^2-x^2+2005x+x-2005+2004\)\(=\left(x-2005\right)x^{2004}-\left(x-2005\right)x^{2003}+\left(x-2005\right)x^{2002}-....+\left(x-2005\right)x^2-\left(x-2005\right)x+\left(x-2005\right)+2004\)\(=\left(x-2005\right)\left(x^{2004}-x^{2003}+x^{2002}-......+x^2-x+1\right)+2004\)
Với x = 2005 => x - 2005 =0
=> A =2004
f(2005)= 2005-2006.2005+2006.2005-2006.2005+2006.2005-2006.2005+2009
f(2005)=4 nha!