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ta có: \(3000^{2009}-1=\left(3000-1\right).\left(3000^{2008}+3000^{2007}+...+3000+1\right)\)
\(=2009.\left(3000^{2008}+3000^{2007}+...+3000+1\right)⋮2009\)
\(\Rightarrow3000^{2009}-1⋮2009\left(đpcm\right)\)
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Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010)
=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010
=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0
=> -2010a + 2.2009b = 0
=> 2010a = 2.2009b
Đề sai
Ta có: (a + 2009)(b - 2010) = (a - 2009)(b + 2010)
=> ab - 2010a + 2009b - 2009.2010 = ab + 2010a - 2009b - 2009.2010
=> ab - 2010a + 2009b - 2009.2010 - ab - 2010a + 2009b + 2009.2010 = 0
=> -2010a + 2.2009b = 0
=> 2010a = 2.2009b Đề sai
Cho: (2010c-2011b)/2009= (2011a-2009c)/2010= (2009b-2010a)/2011
Chứng minh rằng: a/2009=b/2010=c/2011
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\(2018\equiv-1\left(mod2019\right)\)
\(\Rightarrow2018^{2019}\equiv-1^{2019}=-1\) (mod 2019)
\(\Rightarrow2018^{2019}\equiv-1\) (mod 2019)
\(\Rightarrow2018^{2018}+1⋮2019\)
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\(\Leftrightarrow\left(a+2009\right)\left(b-2010\right)=\left(a-2009\right)\left(b+2010\right)\)
=>ab-2010a+2009b-2009x2010=ab+2010a-2009b-2009x2010
=>-4020a=-4018b
=>a/2009=b/2010
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(2008^{100}+2008^{99}\)
\(=2008^{99}.\left(2008+1\right)\)
\(=2008^{99}.2009⋮2009\)
=> đpcm
Học tốt
Bài làm :
Ta có :
\(2008^{100}+2008^{99}=2008^{99}.\left(2008+1\right)=2008^{99}.2009⋮2009\)
=> Điều phải chứng minh