\(\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{2009\cdot2011}+\frac{1}{x}=\frac{1}{5}\cdot0,5\)

\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{2011-2009}{2009\cdot2011}+\frac{1}{x}=\frac{1}{10}\)

\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2009}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)

\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)

\(=\left(\frac{1}{2}\cdot\frac{2006}{10055}\right)+\frac{1}{x}=\frac{1}{10}\)

\(=\frac{1003}{10055}+\frac{1}{x}=\frac{1}{10}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{10}-\frac{1003}{10055}\)

\(\frac{1}{x}=\frac{1}{4022}\)

\(\Rightarrow x=1\div\frac{1}{4022}=4022\)

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\frac{10}{11}\)

\(=\frac{5}{11}\)

\(=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{9\times11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{11}\right)\)

\(=\frac{1}{2}\times\frac{10}{11}\)

\(=\frac{5}{11}\)

           p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)

<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019

<=> p = 1/3 - 1/2019

<=> p = 224/673

\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)

\(=\frac{112}{673}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{2015.2017}+\frac{2}{2017.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2019}\right)=\frac{1}{2}.\frac{224}{673}=\frac{112}{673}\)

\(S.2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(S.2=\frac{1}{1}-\frac{1}{11}\)

\(S.2=\frac{10}{11}\)

\(S=\frac{10}{11}:2\)

\(S=\frac{5}{11}\)

S = 5/11

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=\frac{1}{1}-\frac{1}{11}\)

\(=\frac{10}{11}\)

A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))

A =  \(\dfrac{1}{2}\) \(\times\)  \(\dfrac{2008}{6033}\)

A = \(\dfrac{1004}{6033}\)

\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)

Đến đây bn tự tính nhé.

\(\text {#DNamNgV}\)

\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+...+\dfrac{1}{2009\times2011}\)

\(=\dfrac{1}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{2009\times2011}\right)\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\times\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{2011}\right]\)

\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2008}{6033}\)

\(=\dfrac{1004}{6033}\)

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(x=\frac{9}{7}\div\frac{1}{7}\)

\(x=9\)

Vậy ...

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

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

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

\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)

\(\Rightarrow S=\frac{11}{20}\)

ko bao giờ 323445465

Tớ không chép lại đề nữa nhé:

=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{2009.2011}\right)\)=\(\frac{1}{2}.\left(\frac{3-1}{1-3}+\frac{7-5}{5-7}+...+\frac{2011-2009}{2009-2011}\right)\)

= \(\frac{1}{2}.\left(\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+...+\frac{2011}{2009.2011}-\frac{2009}{2009.2011}\right)\)

=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2009}-\frac{1}{2011}\right)\)

=\(\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

=\(\frac{1}{2}.\frac{2010}{2011}\)

=\(\frac{1005}{2011}\)

bạn ơi đó là dấu nhân hay chữ ''x'' vậy?

nếu là dấu nhân thì như sau:

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{2009\cdot2011}=1\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{5}+...+\frac{1}{2009}\cdot\frac{1}{2011}=1\cdot\frac{1}{2011}=\frac{1}{2011}\)

CHẮC LÀ ĐÚNG ĐÓ BẠN. NHỚ K VÀ KB VS MK NHA. CHÚC BẠN HỌC TỐT. -_-

\(A=\frac{1}{3}-\frac{1}{17}=\frac{14}{51}\)

cách làm thì tự biết

trên mạng đầy

kết quả đúng phải là 7/51 chứ bn 

mk cần cách trình bày thôi 

 câu trả lời của bn hơi lạnh nhạt tí ^.^

\(A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{15.17}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{17}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{17}\right)\)

\(A=\frac{1}{2}.\frac{14}{51}\)

\(A=\frac{7}{51}\)