Câu hỏi của Quan

Question

@Thv · Accepted Answer

$e,\left(7^{2009}-7^{2007}\right):\left(7.7^{2006}\right)\
=\left(7^{2009}-7^{2007}\right):\left(7^{2006+1}\right)\
=\left(7^{2009}-7^{2007}\right):\left(7^{2007}\right)\
=\left(7^{2009}:7^{2007}\right)-\left(7^{2007}:7^{2007}\right)\
=7^{2009-2007}-7^{2007-2007}\
=7^2-7^0\
=49-1\
=48$Câu trả lời: $e,\left(7^{2009}-7^{2007}\right):\left(7.7^{2006}\right)\
=\left(7^{2009}-7^{2007}\right):\left(7^{2006+1}\right)\
=\left(7^{2009}-7^{2007}\right):\left(7^{2007}\right)\
=\left(7^{2009}:7^{2007}\right)-\left(7^{2007}:7^{2007}\right)\
=7^{2009-2007}-7^{2007-2007}\
=7^2-7^0\
=49-1\
=48$

Van Toan · Answer

e,$\left(7^{2009}-7^{2007}\right):\left(7.7^{2006}\right)\
=\left(7^{2009}-7^{2007}\right):7^{2007}\
=7^{2009}:7^{2007}-7^{2007}:7^{2007}\
=7^2-7^0\
=49-1\
=48$Câu trả lời: e,$\left(7^{2009}-7^{2007}\right):\left(7.7^{2006}\right)\
=\left(7^{2009}-7^{2007}\right):7^{2007}\
=7^{2009}:7^{2007}-7^{2007}:7^{2007}\
=7^2-7^0\
=49-1\
=48$

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