\(17.201-7.201-2009=\left(17-7\right).201-2009\)

                                                     \(=10.201-2009\)

                                                     \(=2010-2009=1\)

\(73.\left(-12\right)+\left(-73\right).88=73.\left(-12+\left(-88\right)\right)\)

                                                     \(=73.\left(-100\right)=-7300\)

Học tốt 

So sánh hai phân số : 

\(\frac{2009.2009+2008}{2009.2009+2009}\)và \(\frac{2009.2009+2009}{2009.2009+2010}\)

Vì 20009 x 2009 + 2008 < 2009 x 2009 + 2009

=>A < 1

Ta có: \(B=\frac{2009x2009+2009}{2008x2009+2010}=\frac{2009x\left(2008+1\right)+2009}{2008x2009+2010}=\frac{2008x2009+2009+2009}{2008x2009+2010}\)

\(B=\frac{2008x2009+4018}{2008x2009+2010}=\frac{2008x2009+2010+2008}{2008x2009+2010}=\frac{2008x2009+2010}{2008x2009+2010}+\frac{2008}{2008x2009+2010}\)

\(B=1+\frac{2008}{2008x2009+2010}>1\)

Mà A < 1

=>A < B

dòng đầu sửa chỗ 20009 nhé,2009

- Uầy:)) Mới đây 3 li-ke:)) Nhanh vcc

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

\(2008-\dfrac{2009}{3}-\dfrac{2009}{6}-...-\dfrac{2009}{45}\\ =2008-2009\left(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{45}\right)\\ =2008-2009.2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\\ =2008-4018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)\\ =2008-4018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =2008-4018\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\\ =2008-4018.\dfrac{2}{5}=2008-1607,2\\ =400,8\)

a, \(\frac{1}{2009}+\frac{2}{2009}+...+\frac{2008}{2009}\\ \frac{\left(1+2008\right)\cdot2008\div2}{2009}=\frac{2017036}{2009}\)

Giải:

Ta có:

A=2009²⁰⁰⁸+1/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+2009/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1+2008/2009²⁰⁰⁹+1

2009A=2009²⁰⁰⁹+1/2009²⁰⁰⁹+1 + 2008/2009²⁰⁰⁹+1

2009A=1+2008/2009²⁰⁰⁹+1

Tương tự:

B=2009²⁰⁰⁹+1/2009²⁰¹⁰+1

2009B=1+2008/2009²⁰¹⁰+1

Vì 2008/2009²⁰⁰⁹+1 > 2008/2009²⁰¹⁰+1 nên 2009A>2009B

⇒A>B