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(72005 + 72004) : 72004
= 72005 : 72004 + 72004 : 72004
= 72005 - 2004 + 1
= 71 + 1
= 7 + 1
= 8
~CHÚC BẠN HỌC TỐT~
Ta có : \(\left(7^{2005}+7^{2004}\right):7^{2004}\)
\(=7^{2004}\left(7+1\right):7^{2004}\)
\(=7+1=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(7^{2005}+7^{2004}\right):7^{2004}\)
\(=7^{2005}:7^{2004}+7^{2004}:7^{2004}\)
\(=7^{2005-2004}+1\)
\(=7^1+1\)
\(=7+1=8\)
Tíck cho mìk vs nhoa CHIẾN BINH HẠNH PHÚC!
(72005 + 72004) : 72004
= [72004.(7 + 1)] : 72004
= 7 + 1
= 8
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N=2003(2004(9+8+7+...+2)+2015)+1
Dat A=9+8+7+...+2
A có số số hạng là (9-2)*1+1=8 so hang
A=(9+2)*8/2=44
N=2003(2004*44+2005)+1
N=2003*(88176+2005)+1
N=2003*90181+1=180632543+1=180632544
số to quá
DUYỆT NHA
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7^2005 + 7^2004 - 7^2003
= 7^2002 . 7^3 + 7^2002 . 7^2 - 7^2002 . 7^1
= 7^2002 . ( 7^3 + 7^2 - 7^1 ) chia hết cho 7^2002
Vậy 7^2005 + 7^2004 - 7^2003 chia hết cho 7^2002
Ta có:
\(\left(7^{2005}+7^{2004}-7^{2003}\right)\div7^{2002}\)
\(=\left(7^{2002}\cdot7^3+7^{2002}\cdot7^2-7^{2002}\cdot7\right)\div7^{2002}\)
\(=7^{2002}\left(7^3+7^2-7\right)\div7^{2002}\)
\(=7^3+7^2-7\)
\(=343+49-7\)
\(=385\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(7^{2005}+7^{2004}-7^{2003}\right):7^{2002}\)
\(=7^{2003}\left(7^2+7-1\right):7^{2002}\)
\(=7\left(49+7-1\right)\)
\(=385\)
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\(\left(7^{2005}+7^{2004}\right):7^{2004}=7^{2005}:7^{2004}+7^{2004}:7^{2004}=7+1=8\)
\(\left(11^{2003}+11^{2002}\right):11^{2002}-11^{2003}:11^{2002}+11^{2002}:11^{2002}=11+1=12\)
`(7^2005 + 7^2004) : 7^2004`
`= 7^2005:7^2004 + 7^2004:7^2004`
`= 7+1`
`= 8`
(7^2005 + 7^2004) : 7^2004
=7^2005:7^2004 + 7^2004:7^2004
=7+1
=8