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

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

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

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

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

-1

mình ko chắc đâu đó nha,bài này mình chỉ làm có mấy lần à,sai thì cho mình xin lỗi nhé T_T

Tử = 1+2+2^2+2^3+...+2^2008 
2Tử = 2+2^2+2^3+...+2^2009 
=> 2Tử-Tử=2^2009-1 
S= (2^2009-1)/(1-2^2009)=-1

 

1.

\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)

cứ làm như vậy ta được :

\(=1+1=2\)

2. Ta có :

\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)

vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)

=>x+2011=0

hay x=-2011

Bài làm:

\(A=1-2+3-4+5-...-2008+2009\)

\(A=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2007-2008\right)+2009\)

\(A=-1-1-1-...-1+2009\)(1004 số -1)

\(A=-1004+2009=1005\)

\(B=1+2-3-4+5+6-7-...-2007-2008+2009+2010\)

\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(2006-2007-2008+2009\right)+2010\)

\(B=1+0+0+...+0+2010\)

\(B=2011\)

Học tốt!!!!

ta có: \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1}=\dfrac{2008^{2009}-1+3}{2008^{2009}-1}=1+\dfrac{3}{2008^{2009}-1}\)

B=\(\dfrac{2008^{2009}}{2008^{2009}-3}=\dfrac{2008^{2009}-3+3}{2008^{2009}-3}=1+\dfrac{3}{2008^{2009}-3}\)

ta thấy: \(1+\dfrac{3}{2008^{2009}-1}\)<\(1+\dfrac{3}{2008^{2009}-3}\)

vậy A<B