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a) Ta có:
\(2008^{100}+2008^{99}\)
\(=2008^{99}.\left(2008+1\right)\)
\(=2008^{99}.2009\)
Vì \(2009⋮2009\) nên \(2008^{99}.2009⋮2009.\)
\(\Rightarrow2008^{100}+2008^{99}⋮2009.\)
b) Ta có:
\(12345^{678}-12345^{677}\)
\(=12345^{677}.\left(12345-1\right)\)
\(=12345^{677}.12344\)
Vì \(12344⋮12344\) nên \(12345^{677}.12344⋮12344.\)
\(\Rightarrow12345^{678}-12345^{677}⋮12344.\)
Chúc bạn học tốt!
ta có: 2008100 + 200899 = 200899.(2008+1) = 200899.2009 chia hết cho 2009
=> 2008100 + 200899 chia hết cho 2009 ( đ p c m)
ta có: 12345678 -12345677 = 12345677.(12345-1) = 12345677.12344 chia hết cho 12344
=> đ p c m
\(2008^{100}+2008^{99}=2008^{99}.\left(2008+1\right)=2008^{99}.2009\)
Mà \(2009⋮2009\Rightarrow2008^{99}.2009⋮2009\)
Vậy \(2008^{100}+2008^{99}\)chia hết cho 2009 ( đpcm )
\(12345^{678}-12345^{677}=12345^{677}.\left(12345-1\right)=12345^{677}.12344\)
Mà \(12344⋮12344\Rightarrow12345^{677}.12344⋮12344\)
Vậy \(12345^{678}-12345^{677}\)chia hết cho 12344 ( đpcm )
1. a) Ta thấy: 230=23.10=(23)10=810
320=32.10=(32)10=910
810<910 => 230<320
b) 5202=52.101=(52)101=25101
2505=25.101=(25)101=32101
Mà 25101<32101 =. 5202<2505
2. a) Ta có: 2008100+200899=200899.2008+200899=200899.(2008+1)=200899.2009
200899.2009 chia hết cho 2009 => 2008100+200899 chia hết cho 2009.
b) 12345678-12345677=12345677.12345 - 12345677=12345677.(12345-1)=12345677.12344
=> 12345677.12344 chia hết cho 12344 =. 12345678-12345677 chia hết cho 12344.
k mình nha.
1. a) Ta thấy: 2 30=2 3.10=(2 3 )10=8 10 3 20=3 2.10=(3 2 )10=9 10 8 10<9 10
=> 2 30<3 20 b) 5 202=5 2.101=(5 2 )101=25 101 2 505=2 5.101=(2 5 )101=32 101
Mà 25 101<32 101 =. 5 202<2 505 2. a) Ta có: 2008 100+2008 99=2008 99 .2008+2008 99=2008 99 .(2008+1)=2008 99 .2009
2008 99 .2009 chia hết cho 2009
=> 2008 100+2008 99 chia hết cho 2009.
b) 12345 678 -12345 677=12345 677 .12345 - 12345 677=12345 677 .(12345-1)=12345 677 .12344
=> 12345 677 .12344 chia hết cho 12344 =. 12345 678 -12345 677 chia hết cho 12344.
k mình nha.
Ta có :
\(2008^{100}+2008^{99}\)
\(=2008^{99}.\left(2008+1\right)\)
\(=2008^{99}.2009⋮2009\)
=> đpcm
Học tốt
\(2008^{100}+2008^{99}=2008^{99}.\left(2008+1\right)=2008^{99}.2009⋮2009\) ( đpcm )
\(2008^{100}+2008^{99}\)
\(=2008^{99}.\left(2008+1\right)\)
\(=2008^{99}.2009⋮2009\)
=> đpcm
a) Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2009^{2009}+1+2007}\)
\(A< \frac{2008^{2008}+2008}{2008^{2009}+2008}\)
\(A< \frac{2008.\left(2008^{2007}+1\right)}{2008.\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)
=> A < B
b) Áp dụng \(\frac{a}{b}>1\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\) (a;b;m \(\in\) N*)
Ta có:
\(N=\frac{100^{101}+1}{100^{100}+1}>\frac{100^{101}+1+99}{100^{100}+1+99}\)
\(N>\frac{100^{101}+100}{100^{100}+100}\)
\(N>\frac{100.\left(100^{100}+1\right)}{100.\left(100^{99}+1\right)}=\frac{100^{100}+1}{100^{99}+1}=M\)
=> M > N
\(2008^{100}\cdot2008^{99}\)
\(=2008^{99}\left(2008+1\right)\)
\(=2008^{99}\cdot2009⋮2009\left(đpcm\right)\)
\(2^{50}=\left(2^5\right)^{10}=32^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Suy ra: 250 > 520
b)
\(9^{200}=\left(9^2\right)^{100}=81^{100}\)
Suy ra: 99100 > 81100
a, 2008\(-⋮\)-1(mod 2009)
\(2008^{100}-⋮1\left(mod2009\right)\)
\(2008^{99}-⋮-1\left(mod2009\right)\)
=>\(2008^{100}+2008^{99}⋮2009\)
b,\(12345-⋮1\left(mod12344\right)\)
\(12345^{678}-⋮1\left(mod12344\right)\)
\(12345^{677}-⋮1\left(mod12344\right)\)
\(12345^{678}+12345^{677}không⋮12344\)(đề sai)
\(-⋮\)là đồng dư nha