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31 tháng 7 2016

\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

   \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)

   \(=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)

   \(=\frac{1}{2}.\frac{48}{49}\)

   \(=\frac{24}{49}\)

31 tháng 7 2016

\(K\times2=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\)

\(K\times2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\)

\(K\times2=\frac{48}{49}\)

\(K=\frac{48}{49}\div2=\frac{24}{49}\)

31 tháng 7 2016

\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

    \(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\right):2\)

     =  \(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right):2\)   

     = \(\left(1-\frac{1}{49}\right):2\)

     \(=\frac{48}{49}:2\) \(\frac{24}{49}\)

 

31 tháng 7 2016

\(\frac{48}{49}\)

20 tháng 8 2016

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)

9 tháng 4 2018

\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(100=2x+4\)

\(\Leftrightarrow\)\(2x=96\)

\(\Leftrightarrow\)\(48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

9 tháng 4 2018

\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)

\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(49=x+1\)

\(\Leftrightarrow\)\(x=48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

21 tháng 8 2016

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\), ta có:

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

21 tháng 8 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

mk đầu tiên đấy

19 tháng 3 2019

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2003.2005}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2003}-\frac{1}{2005}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2005}\right)=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}=\)

\(=\frac{2}{2.1.3}+\frac{2}{2.3.5}+\frac{2}{2.5.7}+....+\frac{2}{2.2003.2005}\)

\(=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2003.2005}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)

Chúc bạn học tốt nha!

9 tháng 5 2017

\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\)

2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{7.9}\)

2A = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)

2A = \(1-\frac{1}{9}=\frac{8}{9}\)

A = \(\frac{8}{9}:2=\frac{4}{9}\)

Đặt B = \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\)

2B = \(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{8.10}\)

2B = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)

2B = \(\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

B = \(\frac{2}{5}:2=\frac{1}{5}\)

Thay A và B vào S ta được:

\(S=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)

9 tháng 5 2017

\(S=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{7.9}+\frac{1}{8.10}\)

\(\Rightarrow S=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)

\(S=\frac{1}{2}.\frac{8}{9}+\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{1}{2}\left(\frac{8}{9}+\frac{2}{5}\right)\)

\(S=\frac{1}{2}.\frac{58}{45}\)

\(S=\frac{29}{45}\)

23 tháng 3 2020

\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)

=> \(2P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{29.31}\)

\(=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{31-29}{29.31}\)

\(=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{29}-\frac{1}{31}\right)\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

=> \(P=\frac{30}{31}:2=\frac{15}{31}\)

23 tháng 3 2020

Nếu đề là tính thì bạn làm như sau nhé :

\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)

\(\Rightarrow2P=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{29.31}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow P=\frac{30}{31}\div2=\frac{15}{31}\)

4 tháng 5 2016

 nhung ma ko cothoi gian giai

4 tháng 5 2016

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)