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

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

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

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

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

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

\(=\dfrac{2^{2009}-1}{1-2^{2009}}=-1\)

-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

 

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!!!!

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

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

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

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

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