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A=(2009+2009^2)+(2009^3+2009^4)+...+(2009^9+2009^10)
A=[2009.(1+2009)]+[2009^3.(1+2009)]+....+[2009^9.(1+2009)]
A=2009.2010+2009^3.2010+...+2009^9.2010
A=2010(2009+2009^3+2009^5+......+2009^9) chia het cho 2010
Ta có :
\(A=2009+2009^2+2009^3+2009^4+....+2009^{10}\)
Tổng A có số số hạng là :
( 10 - 1 ) : 1 + 1 = 10 ( số hạng )
Vì \(10⋮2\)nên khi ta nhóm 2 số liên tiếp lại thành một căp thì không thừa số nào cả
\(\Rightarrow A=\left(2009+2009^2\right)+\left(2009^3+2009^4\right)+....+\left(2009^9+2009^{10}\right)\)
\(\Rightarrow A=2009.\left(1+2009\right)+2009^3.\left(1+2009\right)+....+2009^9.\left(1+2009\right)\)
\(\Rightarrow A=2009.2010+2009^3.2010+....+2009^9.2010\)
\(\Rightarrow A=2010.\left(2009+2009^3+....+2009^9\right)\)
Vì \(2009+2009^3+....+2009^9\inℤ\)nên \(2010.\left(2009+2009^3+....+2009^9\right)\inℤ\)
Vì \(2010⋮2010\)nên \(A⋮2010\)
Vậy \(A=2009+2009^2+2009^3+....+2009^{10}⋮2010\left(ĐPCM\right)\)
Đặt \(A=\frac{2009^{2008}+1}{2009^{2009}+1}\)và \(B=\frac{2009^{2009}+1}{2009^{2010}+1}\)
\(A=\frac{2009^{2008}+1}{2009^{2009}+1}\Rightarrow2009A=\frac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\frac{2009^{2009}+2009}{2009^{2009}+1}=1+\frac{2008}{2009^{2009}+1}\)
\(B=\frac{2009^{2009}+1}{2009^{2010}+1}\Rightarrow2009B=\frac{2009.\left(2009^{2009}+1\right)}{2009^{2010}+1}=\frac{2009^{2010}+2009}{2009^{2010}+1}=1+\frac{2008}{2009^{2010}+1}\)
Vì \(\frac{2008}{2009^{2009}+1}>\frac{2008}{2009^{2010}+1}\Rightarrow2009A>2009B\Rightarrow A>B\)
1)Đặt A=1+2+22+23+.....+22008
=>2A=2+22+23+....+22009
=>2A-A=(2+22+23+...+22009)-(1+2+22+23+....+22008)
=-1+22009
\(2009A=\frac{2009^{2010}+2009}{2009^{2010}+1}=\)\(\frac{2009^{2010}+1+2008}{2009^{2010}+1}=1+\frac{2008}{2009^{2010}+1}\)
\(2009B=\frac{2009^{2009}+2009}{2009^{2009}+1}=\frac{2009^{2009}+1+2008}{2009^{2009}+1}\)\(=1+\frac{2008}{2009^{2009}+1}\)
Vì \(1+\frac{2008}{2009^{2010}+1}< 1+\frac{2008}{2009^{2009}+1}\) \(\Leftrightarrow A< B\)
\(A=\frac{2009^{2009}+1}{2009^{2010}+1}\Rightarrow2009A=\frac{2009^{2010}+2009}{2009^{2010}+1}\)
\(2009A=\frac{2009^{2010}+1}{2009^{2010}+1}+\frac{2008}{2009^{2010}+1}\)
\(2009A=1+\frac{2008}{2009^{2010}+1}\)
..... sory bn mk hơi luwoif chút nên bn tự lm tương tự vs phần B và so sánh nhé!^^
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Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:
\(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)
\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)
\(=2009^{2008}-1\)
\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)
\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008
=> ĐPCM
Chứng Minh Rằng: (2009+20092+20093+20094+...+20092009)-(1+2009+20092+20093+...+20092008) chia hết cho 2008.
Đặt A=2009+20092+20093+20094+...+20092009, B=1+2009+20092+20093+20094+...+20092008
Ta có:
+)A=2009+20092+20093+20094+...+20092009
2009A= 20092+20093+20094+...+20092010
2009A-A=(20092+20093+20094+...+20092010)-(2009+20092+20093+20094+...+20092009)
2008A=20092010- 2009
=> A=(20092010- 2009)/2008
=> A chia hết cho 2008.
B=1+2009+20092+20093+20094+...+20092008
2009B=2009+20092+20093+20094+...+20092010
2009B-B=(2009+20092+20093+20094+...+20092010)-(1+2009+20092+20093+20094+...+20092009)
2008B=20092010-1
=>B=(20092010-1)/2008
=>B chia hết cho 2008
=> A-B chia hết cho 2008.
=> ĐPCM