Ta có:

$S=\dfrac{5}{1\cdot2\cdot3}+\dfrac{8}{2\cdot3\cdot4}+\dfrac{11}{3\cdot4\cdot5}+\cdots+\dfrac{6026}{2008\cdot2009\cdot2010}$

Nhận thấy tử số có dạng $3n+2$, nên:

$S=\sum_{n=1}^{2008}\dfrac{3n+2}{n(n+1)(n+2)}$

Ta có: $\dfrac{3n+2}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)}+\dfrac{2}{(n+1)(n+2)}$

Do đó:

$S=\left(\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\cdots+\dfrac1{2008\cdot2009}\right)$

$+2\left(\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{2009\cdot2010}\right)$

Mà: $\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}$

Nên: $S=\left(1-\dfrac1{2009}\right)+2\left(\dfrac12-\dfrac1{2010}\right)$

$=1-\dfrac1{2009}+1-\dfrac1{1005}$

$=2-\dfrac1{2009}-\dfrac1{1005}$

Vì: $\dfrac1{2009}+\dfrac1{1005}>0$

Nên $S<2$

B=\(\frac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)=\(\frac{2+2^2+2^3...+2^{2009}-1-2-2^2-...-2^{2008}}{\left(1-2^{2009}\right)}\)=\(\frac{2^{2009}-1}{1-2^{2009}}\)=-1

Vậy: B=-1

\(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)

\(2B=\frac{2+2^2+2^3+...+2^{2009}}{1-2^{2009}}\)

\(2B-B=\frac{\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2+2^2+2^3+...+2^{2008}\right)}{1-2^{2009}}\)

\(B=\frac{2^{2009}-1}{1-2^{2009}}\)

\(B=-1\)

Lời giải:
Xét tử số:
$X=1+2+2^2+2^3+...+2^{2008}$

$2X=2+2^2+2^3+2^4+....+2^{2009}$

$\Rightarrow 2X-X=(2+2^2+2^3+2^4+....+2^{2009})-(1+2+2^2+...+2^{2008})$

$\Rightarrow X=2^{2009}-1$

$\Rightarrow S=\frac{X}{1-2^{2009}}=\frac{2^{2009}-1}{-(2^{2009}-1)}=-1$

a, 1004 .2009 + 1005 = (1005-1) .2009 +1005

                                = 1005 .2009 -2009 +1005

                                = 1005 .2009 -1004

Vậy ( 1004 .2009 +1005) / (1005 .2009 -1004) =1

b, 1004 .2010 +1 = 1004 .2009 +1004 +1

                          = (1006 -2) .2009 +1005

                          = 1006 .2009 -2 .2009 +1005

                          = 1006 .2009 -4008 +1005

                          = 1006 .2009 -3013

Vậy (1004 .2010 +1) / (1006 .2009 -3013) = 1

c, 2007 .2009 -2 = 2007.(2008+1) -2

                         = 2007.2008 +2007 -2

                         = 2007.2008 +2005

                         = (2008-1) .2008 +2005

                         = 2008 .2008 -2008 +2005

                         = 2008 .2008 -3

Vậy (2008 .2008 -3) / (2007 .2009 -2) =1

bam may đi bạn tui

\(B=2^{2010}-2^{2009}+2^{2008}-...+2^2-2\)

\(2B=2^{2011}-2^{1010}+2^{2009}-...+2^3-2^2\)

\(3A=2^{2011}-2\)

\(A=\frac{2\left(2^{2010}-1\right)}{3}\)

dễ ợt

s=2010(1+20100+2010^3(1+2010)+............+2010^2009(1+2010)

s=2010.2011+2010^3.2011+.........+2010^2009.2011

s=2011(2010+2010^3+.......+2010^2009) chia hết cho 2011

 \(S=\left(2010+2010^2\right)+\left(2010^3+2010^4\right)+...+\left(2010^{2009}+2010^{2010}\right)\)

\(S=2010\left(2010+1\right)+2010^3\left(2010+1\right)+...+2010^{2009}\left(2010+1\right)\)

 \(S=2011.\left(2010+2010^3+2010^5+...+2010^{2009}\right)\) chia hết cho 2011

a.|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=4 hoăc x=-2

vậy...

b.2x+17=15

=>2x=15-17

=>2x=-2

=>x=-1

vậy...

c.tự giải

d.4x-15=-75-x

=>4x+x=-75+15

=>5x=-60

=>x=-12

vậy...

2.2+(-3)+4+...+(-2011)+2012

=(-3+2)+...+(-2011+2010)+2012 (có 2010 cặp)

=-1.2010+2012

=-2010+2012=2

Ai mún kb vs mink ko