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em chưa học ẹ

a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)

\(=\dfrac{2}{5}+\dfrac{4}{2}\)

\(=\dfrac{2}{5}+2\)

\(=\dfrac{2}{5}+\dfrac{10}{5}\)

\(=\dfrac{12}{5}\)

b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)

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

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

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

\(=1-\dfrac{2}{2008}\)

\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)

\(=\dfrac{2006}{2008}\)

\(=\dfrac{1003}{1004}\)

a: =2/5+4/2

=2/5+2

=12/5

b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

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

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

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

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

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