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a.2014100 + 201499
=201499.(2014+1)
=201499.2015
=> 2014100 + 201499 chia hết cho 2015
b.31994 + 31993 _ 31992
=31992.(32+3-1)
=31992.11
=>31994 + 31993 _ 31992 chia hết cho 11
c. 413 _ 325 _ 88
=(22)13-(25)5-(23)8
=226-225-224
=224.(22-2-1)
=224.5
=> 413 _ 325 _ 88 chia hết cho 5
a)\(2014^{100}+2014^{99}=2014^{99}.\left(2014+1\right)=2014^{99}.2015⋮2015\left(\text{Đ}PCM\right)\)
b)\(3^{1994}+3^{1993}-3^{1992}=3^{1992}.\left(3^2+3-1\right)=3^{1992}.\left(9+3-1\right)=3^{1992}.11⋮11\left(\text{Đ}PCM\right)\)
c)\(4^{13}-32^5-8^8=\left(2^2\right)^{13}-\left(2^5\right)^5-\left(2^3\right)^8=2^{26}-2^{25}-2^{24}=2^{24}.\left(2^2-2-1\right)\)
Đề sai rồi bạn 2^14 luôn tận cùng chẵn =>2^14 không chia hết cho 5
Chúc bạn học tốt
a.
\(\left(0,25\right)^3\times32\)
\(=\left(0,25\right)^3\times2^5\)
\(=\left(0,25\right)^3\times2^3\times2^2\)
\(=\left(0,25\times2\right)^3\times4\)
\(=\left(0,5\right)^3\times4\)
\(=0,125\times4\)
\(=0,5\)
b.
\(\left(-0,125\right)^3\times80^4\)
\(=\left(-0,125\right)^3\times80^3\times80\)
\(=\left(-0,125\times80\right)^3\times80\)
\(=\left(-10\right)^3\times80\)
\(=-1000\times80\)
\(=-80000\)
c.
\(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}\times\left(3^2+3-1\right)\)
\(=3^{1992}\times\left(9+3-1\right)\)
\(=3^{1992}\times11\)
\(\Rightarrow3^{1994}+3^{1993}-3^{1992}⋮11\)
d.
\(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\times\left(2^2+2-1\right)\)
\(=2^{24}\times\left(4+2-1\right)\)
\(=2^{24}\times5\)
\(\Rightarrow4^{13}+32^5-8^8⋮5\)
Chúc bạn học tốt
a) \(2010^{100}\)+ \(2010^{99}\)
= \(2010^{99}\)\(\left(2010+1\right)\)
= \(2010^{99}\). \(2011\)chia hết cho 2011
Vậy ...................................
b) \(3^{1994}\)+ \(3^{1993}\)- \(3^{1992}\)
= \(3^{1992}\)\(\left(3^2+3-1\right)\)
= \(3^{1992}\). \(11\)
Vậy .......................
c) \(4^{13}\)+ \(32^5\)- \(8^8\)
= \(\left(2^2\right)^{13}\)+ \(\left(2^5\right)^5\)- \(\left(2^3\right)^8\)
= \(2^{26}\)- \(2^{25}\)- \(2^{24}\)
= \(2^{24}\). \(\left(2^2+2-1\right)\)
= \(2^{24}\). \(5\)
Vậy .......................
3 cau 3 nhe
a)
\(=2010^{99}\left(2010+1\right)\)
\(=2010^{99}.2011\)
cung thay chia het ro nhi
b)
\(=3^{1992}\left(3^2+3-1\right)\)
\(=3^{1992}.11\)
cung thay chia het ro nhi
c)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5\)
cung thay chia het ro nhi
cho 3 nhe
Có: \(3^{1994}+3^{1993}-3^{1992}=3^{1992}\left(3^2+3-1\right)=11\cdot3^{1992}\)
=>đpcm
1) 3^1994+4^1993-3^1992
= 3^1992.(9+3-1)=3^1992.11 chia hết cho 11
=> 3^1994+3^1993-3^1992 chia hết cho 11
a) \(2010^{100}+2010^{99}\)
\(=2010^{99}\left(2010+1\right)\)
\(=2010^{99}.2011⋮2011\left(dpcm\right)\)
b) \(3^{1994}+3^{1993}-3^{1992}\)
\(=3^{1992}\left(3^2+3-1\right)\)
\(=3^{1992}.11⋮11\left(dpcm\right)\)
c) \(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5⋮5\left(dpcm\right)\)