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

29,83x73+29,83x26+29,83

=29,83x(73+26+1)

=29,83x 100

=3983

200-198+196-194+192-190+.............+8-6+4-2

=(200+196+192+...+8+4)   -     ( 198+194+190+...+6+2)

= có 50 số hạng                                có 50 số hạng

=(200+4):2x50                   -         (198+2):2x50

=5100                               -       5000

=          100

\(2010^2-2009^2+2008^2-...+2^2-1^2\)

\(=-\left(1^2-2^2+3^2-...+2009^2-2010^2\right)\)

\(=-\left[1^2+2^2+...+2009^2+2010^2-\left(2^2+4^2+...+2010^2\right)\right]\)

\(=-\left[\frac{2010.\left(2010-1\right)\left(2.2010-1\right)}{6}-2^2\left(1^2+2^2+...+1005^2\right)\right]\)

\(=-\left[2704847285-2^2.\frac{1005\left(1005-1\right)\left(2.1005-1\right)}{6}\right]\)

\(=-\left(2704847285-1351414120\right)=1353433165\)

 2010×2010 - 2009×2009 +2008×2008-...+2×2-1×1

=2 x 2010 - 2 x 2009 + .......+ 2 x 2 - 2 x 1

=2x(2010-2009+2008-.......+2-1)

=2x[(2010-2019)+......+(2-1)]

=2x ( 1+ 1+....+1)

=2x1005

=2010

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