កំណែលំហាត់សិស្សពូកែភ្នំពេញ ២០១១ (៨) ៖ គណនាតម្លៃនៃ A

February 26, 2011 By វិចិត្រ Leave a Comment

ក) រកតម្លៃនៃ $A=2011^3-2010.2011^2-2010^2.2011+2010^3$
ខ) រកតម្លៃនៃ $B=403^5-402^2 \left(403^3+2.403^2+3.403+4\right)$

ចម្លើយ

ក) យើងមាន

$A=2011^3-2010.2011^2-2010^2.2011+2010^3$
$=2011^2 (2011-2010)+2010^2 (2010-2011)$
$=2011^2-2010^2$
$=(2011-2010)(2011+2010)$
$=4021$

ខ) យើងមាន

$B=403^5-402^2 \left(403^3+2.403^2+3.403+4\right)$
$=403^5-(403-1)^2\left(403^3+2.403^2+3.403+4\right)$
$=403^5-\left(403^2-2.403+1\right)\left(403^3+2.403^2+3.403+4\right)$
$=-(-5.403+4)$
$=2011$

Be Sociable, Share!

Tweet

Related Posts

Filed Under: គណិតវិទ្យា, អប់រំ Tagged With: algebra, olympiad, គណិតវិទ្យា