a=2009(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.........+1/2008-1/2009)

=2009x2008/2009

=2008

em chưa học ẹ

b) Giải:

Ta có: \(k\left(k+1\right)\left(k+2\right)\)

\(=\dfrac{1}{4}\left[k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\left(k-1\right)k\left(k+1\right)\left(k+2\right)\right]\)

Do đó: \(P=\dfrac{1}{4}.n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)

Thay vào ta tính được:

\(P\left(100\right)=26527650;P\left(2009\right)=\dfrac{1}{4}.2009.2010.2011.2012\)

Mà: \(\dfrac{1}{4}.2009.2010.2011=2030149748\)

Và \(149748.2012=3011731776;2030.2012.10^6=4084360000000\)

Cộng lại ta có: \(P\left(2009\right)=4087371731776\)

Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010}\)

\(\Rightarrow x+1=2010\)

\(\Rightarrow x=2010-1\)

\(\Rightarrow x=2009\)

Vậy x = 2009

=> 1-1/2+1/2-1/3+1/3- 1/4 +... +1/x -1/x+1 = 2009/1020

=> 1 - 1/x+1=2009/2010

=> (x+1-1)/x+1=2009/2010

=> x/x+1=2009/2010

=>x=2009

phúc hơi phức tạp

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\frac{1}{x+1}=1-\frac{2008}{2009}\)

\(\frac{1}{x+1}=\frac{1}{2009}\)

\(\Rightarrow x+1=2009\)

\(x=2009-1\)

\(x=2008\)

Vậy \(x=2008\)

Tự làm bước biến đổi nhé tui lm lẹ luôn =v

\(\frac{1}{1}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\frac{x}{x+1}=\frac{2008}{2009}\)

\(=>x=2008\)

Vậy x = 2008

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2008}{2009}\Leftrightarrow\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\frac{x}{x+1}=\frac{2008}{2009}\Leftrightarrow2009x=2008x+2008\Leftrightarrow x=2008\)

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