A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)

2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)

2A= 1-\(\dfrac{1}{99}\)

2A= \(\dfrac{98}{99}\)

  A= \(\dfrac{98}{99}\) : 2

A=\(\dfrac{49}{99}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)

A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2017.2019

A = 1/2 (1 - 1/3 + 1/3 - 1/5 + 1/5 - ... - 1/2019)

A = 1/2 (1 - 1/2019)

A = 1/2 . 2018/2019

A = 1009/2019

@Cỏ

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{2017\cdot2019}\)

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

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}\cdot\frac{2018}{2019}\)

\(=\frac{1009}{2019}\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

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

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1}{2}.\frac{2018}{2019}=\frac{1009}{2019}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)

\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)

\(2A=1-\frac{1}{10}\)

\(2A=\frac{9}{10}\)

\(A=\frac{9}{10}:2=\frac{9}{20}\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)

( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)

=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)

6Q = 1.3.6 + 3.5.(7-1) + 5.7.(9-3) + ... + 1999.2001.(2003-1997)

6Q = 18 + 3.5.7 - 1.3.5 + 5.7.9 - 3.5.7 + ... + 1999.2001.2003 - 1997.1999.2001

6Q = (18 + 3.5.7 + 5.7.9 + ... + 1999.2001.2003) - (1.3.5 + 3.5.7 + ... + 1997.1999.2001)

6Q = 18 + 1999.2001.2003 - 1.3.5

6Q = 18 + 1999.2001.2003 - 15

6Q = 3 + 8011997997

6Q = 8011998000

Q = 1335333000

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)

=> x + 2 = 41 

=> x = 39

I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)

\(\Leftrightarrow3n+6-2⋮n+2\)

\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{-1;-3;0;-4\right\}\)

II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)

\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)

\(S=\dfrac{1}{2}.\left(\dfrac{2}{\sqrt{1.3}}+\dfrac{2}{\sqrt{3.5}}+.......+\dfrac{2}{\sqrt{29.31}}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}}+.....-\dfrac{1}{\sqrt{29}}+\dfrac{1}{\sqrt{29}}-\dfrac{1}{\sqrt{31}}\right)\)

\(S=\dfrac{1}{2}.\left(1-\dfrac{1}{\sqrt{31}}\right)=\dfrac{1}{2}.\left(\dfrac{31-\sqrt{31}}{31}\right)=\dfrac{31-\sqrt{31}}{62}\)