toán 7

P

pinkylun

Ta có:

$x=2005$=>$2006=x+1$

thay vào ta đc:

$x^{2006} - 2006x ^{2005} + 2006x^{2004} - 2006x^{2003} +...- 2006x^2 +2006x - 1$

$=x^{2006}-(x+1)x^{2005}+...-(x+1)x^{2}+(x+1)x-1$

$=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...-x^3-x^2+x^2+x-1$

$=x-1=2005-1=2004$
 
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