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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) \(4^{13}+32^5-8^8\)
\(=\left(2^2\right)^{13}+\left(2^5\right)^5-\left(2^3\right)^8\)
\(=2^{26}+2^{25}-2^{24}\)
\(=2^{24}\left(2^2+2-1\right)\)
\(=2^{24}.5⋮5\left(dpcm\right)\)
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
Câu 1:
10^19+10^18+10^17
=10^17(10^2+10+1)
=10^17.111
=10^16.10.111
=10^16.1110 chia hết cho 555
suy ra 10^19+10^18+10^17 chia hết cho 555
a) Có: \(4^{51}+2^{104}+4^{53}\\ =4^{51}+\left(2^2\right)^{52}+4^{53}\\ =4^{51}+4^{52}+4^{53}\\ =4^{51}\left(1+4+4^2\right)\\ =4^{51}\cdot21⋮21\left(đpcm\right)\)
b) Có: \(125^{10}+5^{31}+25^{16}\\ =\left(5^3\right)^{10}+5^{31}+\left(5^2\right)^{16}\\ =5^{30}+5^{31}+5^{32}\\ =5^{30}\left(1+5+5^2\right)\\ =5^{30}\cdot31⋮31\left(đpcm\right)\)
c) Có: \(2^{25}+4^{13}+8^9\\ =2^{25}+\left(2^2\right)^{13}+\left(2^3\right)^9\\ =2^{25}+2^{26}+2^{27}\\ =2^{23}\left(2^2+2^3+2^4\right)\\ =2^{23}\cdot28⋮28\left(đpcm\right)\)
A = 75 . ( 41993 + 41992 + ... + 42 + 4 + 1 ) + 25
A = 25 . 3 . ( 41993 + 41992 + ... + 42 + 4 + 1 ) + 25
A = 25 . [ 4 . ( 41993 + 41992 + ... + 42 + 4 + 1 ) - ( 41993 + 41992 + ... + 42 + 4 + 1 ) ] + 25
A = 25 . [ ( 41994 + 41993 + ... + 43 + 42 + 1 ) - ( 41993 + 41992 + ... + 42 + 4 + 1 ) ] + 25
A = 25 . ( 41994 - 1 ) + 25
A = 25 . ( 41994 - 1 + 1 )
A = 25 . 41994
A = 25 . 4 . 41993
A = 100 . 41993 \(⋮\)100
2.
a) gọi 3 số nguyên liên tiếp là a , a + 1 , a + 2
Theo bài ra : a + ( a + 1 ) + ( a + 2 ) = ( a + a + a ) + ( 1 + 2 ) = 3a + 3 = 3 . ( a + 1 ) \(⋮\)3
b) gọi 5 số nguyên liên tiếp là b, b + 1 , b + 2 , b + 3 , b + 4
Theo bài ra : b + ( b + 1 ) + ( b + 2 ) + ( b + 3 ) + ( b + 4 )
= ( b + b + b + b + b ) + ( 1 + 2 + 3 + 4 )
= 5b + 10
= 5 . ( b + 2 ) \(⋮\)5
3.
Ta có : \(\frac{10^{94}+2}{3}=\frac{10...0+2}{3}=\frac{100...002}{3}\text{ }⋮\text{ }3\)là số nguyên
\(\frac{10^{94}+8}{9}=\frac{100...00+8}{9}=\frac{100...008}{9}\text{ }⋮\text{ }9\)là số nguyên
\(8^{30}+8^{31}+8^{32}\)
\(=8^{30}.1+8^{30}.8+8^{30}.8^2\)
\(=8^{30}.1+8^{30}.8+8^{30}.64\)
\(=8^{30}\left(1+8+64\right)\)
\(=8^{30}.73\)
\(=\left(2^3\right)^{30}.73\)
\(=2^{90}.73\)
\(=2^{89}.146⋮146\rightarrowđpcm\)
\(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)
\(=4^{25}.1+4^{25}.4+4^{25}.4^2+4^{25}.4^3+4^{25}.4^4+4^{25}.4^5\)
\(=4^{25}.1+4^{25}.4+4^{25}.16+4^{25}.64+4^{25}.256+4^{25}.1024\)
\(=4^{25}\left(1+4+16+64+256+1024\right)\)
\(=4^{25}.1365\)
\(=4^{25}.195.7⋮7\rightarrowđpcm\)
à há, giờ mới biết mi làm sao biết đc cách giải BTVN