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

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

Ta có  \(A=\dfrac{2}{1.3}-\dfrac{2}{2.4}+\dfrac{2}{3.5}-\dfrac{2}{4.6}+\dfrac{2}{5.7}-\dfrac{2}{6.8}+\dfrac{2}{7.9}-\dfrac{2}{8.10}+\dfrac{2}{9.11}-\dfrac{2}{10.12}\) 

\(\Rightarrow A=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\right)-\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+\dfrac{2}{10.12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}\right)\) \(\Rightarrow A=\left(1-\dfrac{1}{11}\right)-\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\) 

\(\Rightarrow A=1-\dfrac{1}{11}-\dfrac{1}{2}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{9}{22}+\dfrac{1}{12}\) 

\(\Rightarrow A=\dfrac{65}{132}\) 

Mà \(\dfrac{65}{132}< 1\) \(\Rightarrow A< 1\) 

Vậy \(A< 1\)

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

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

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

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

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

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

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

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

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

✫¸.•°*”˜˜”*°•✫ Ṱђầภ Ḉђết ✫•°*”˜˜”*°•.¸✫ nhân A với 2 rồi phân tích như vậy được

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+....+\frac{1}{37\cdot39}\)

\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+....+\frac{2}{37\cdot39}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{37}-\frac{1}{39}\)

\(2A=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)

\(A=\frac{4}{13}:2=\frac{4}{13}\cdot\frac{1}{2}=\frac{2}{13}\)

Vậy \(A=\frac{2}{13}\)

Bài làm

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(A=\frac{1}{3}-\frac{1}{39}\)

\(A=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

Vậy \(A=\frac{12}{39}\)

 A  =1/3.5+1/5.7 +1/7.9 + ... +1/37.39

2.A = 2/3.5+ 2/5.7+ ....+2/37.39

2.A = 1/3- 1/5+1/5-1/7 +....+1/37-1/39

      = 1/3- 1/39

2.A= 13/39 -1/39

2.A= 12/39

A  = 12/39: 2

A= 2/13

Vậy A= 2/13

=>2A=1/3-1/5+1/5-1/7+1/7-1/9+...+1/37-1/39

=>2A=1/3-1/39=4/13

=>A=2/13

  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=\(\dfrac{49}{99}\)

\(A=-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)-\frac{1}{27}\)

\(=-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)-\frac{1}{27}\)

\(=-\left(1-\frac{1}{27}\right)-\frac{1}{27}\)

\(=-1+\frac{1}{27}-\frac{1}{27}\)

\(=-1\)

1/2015

ok, cảm ơn bạn