toán đại 6

S

su10112000a

ta có:
$1102^{2009} -1102^{2008} - 1102^{2008} + 1102^{2007}$
$=1102^{2008}(1102-1) - 1102^{2007}(1102-1)$
$=1102^{2008}.1101-1102^{2007}.1101$
vì $1101=1101$ nên $1102^{2008}.1101 > 1102^{2007}.1101$
 
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