dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A=1-\frac{1}{2010}\)

\(A=\frac{2009}{2010}\)

a) = 1-1/2+1/2-1/3+1/3-1/4

    = 1-1/4=3/4

b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018

   =1-1/2018=2017/2018

c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015

   = 1/2-1/2015=2015/4030-2/4030=2013/4030

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)

b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)

\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)

\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)

\(=\frac{3}{2}.\frac{2013}{4030}\)

\(=\frac{6039}{8060}\)

]\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(K=2\times\frac{502}{1005}\)

\(K=\frac{1004}{1005}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)

\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(3F=\frac{1}{3}-\frac{1}{33}\)

\(F=\frac{10}{33}:3\)

\(F=\frac{10}{99}\)

\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(I=1-\frac{1}{2010}\)

\(I=\frac{2009}{2010}\)

=1-1/2+1/2-1/3+1/3+1/4+.....+1/2009-1/2010

=1-1/2010

=-1/2009

lâu lằm k làm mấy bài kiểu nay... đúng k bạn ơi?????^-^

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

\(=1-\frac{1}{2010}=\frac{2009}{2010}\)

2009/2010

Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(A=1-\frac{1}{200}\)

\(A=\frac{199}{200}\)

\(=1-\frac{1}{200}=\frac{199}{200}\)

\(A=\frac{1}{1.2}+............+\frac{1}{199.200}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...........+\frac{1}{199}-\frac{1}{200}\)

\(=1-\frac{1}{200}\)

\(=\frac{199}{200}\)

Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{a\left(a+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{a}-\frac{1}{a+1}\)

\(=1-\frac{1}{a+1}\)

\(=\frac{a+1}{a+1}-\frac{1}{a+1}=\frac{a}{a+1}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{a\left(a+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}\)

\(=1-\frac{1}{a+1}\)

\(=\frac{a+1}{a+1}-\frac{1}{a+1}\)

\(=\frac{a}{a+1}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{a.\left(a+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{a}-\frac{1}{a+1}\)

\(=1-\frac{1}{a+1}\)

\(=\frac{a}{a+1}\)

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