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

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

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

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

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

\(B=-1\)

\(1-3+3^2-3^3+....-3^{2007}+3^{2008}\)

\(3S=3-3^2+3^3-3^4+...-3^{2008}+3^{2009}\)

\(4S=3^{2009}+1\)

\(\Rightarrow A=4S-1-3^{2009}\)

\(=\left(3^{2009}+1\right)-1-3^{2009}\)

\(=0\)

S=1−2+2^2+2^3+...+2^2000

2S=2−2^2+2^3−2^4+...+2^2001

⇒2S-S=2^2001-1

⇒S=.................................

Ta có : S=1-2+2²-2³+...+2¹⁰⁰⁰ 

=>2S=2-2²+2³-2⁴+...+2¹⁰⁰¹

=>3S=1+2¹⁰⁰¹

=>S=\(\frac{1+2^{1001}}{3}\)

a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)

\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)

\(\Leftrightarrow8x=-\frac{5}{4}\)

\(\Leftrightarrow x=-\frac{5}{32}\)

c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)

\(\Leftrightarrow x+1=2003\)

\(\Leftrightarrow x=2002\)

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

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 $S<2$

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

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