\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{2021.2023}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{2021.2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}.\dfrac{2022}{2023}=\dfrac{1011}{2023}\)

Ta có A = \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{2021\cdot2023}\)

            = \(\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}+\dfrac{1}{2023}\right)\)

            = \(\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)=\dfrac{1}{2}\cdot\dfrac{2022}{2023}=\dfrac{1011}{2023}\)
 

A=1.31​+3.51​+5.71​+.....+2021.20231​

=12.(21.3+23.5+25.7+....+22021.2023)=21​.(1.32​+3.52​+5.72​+....+2021.20232​)

=12.(1−13+13−15+15−17+....+12021−12023)=21​.(1−31​+31​−51​+51​−71​+....+20211​−20231​)

=12.(1−12023)=12.20222023=10112023=21​.(1−20231​)=21​.20232022​=20231011​

mình sẽ ủng hộ bạn có câu trả lời đúng nhất nhé

6F=1.3(5-1)+3.5(7-1)+5.7(9-3)+...99.101(103-97)

6F=1.3+1.3.5-1.3.5+3.5.7-3.5.7+.....-97.99.101+99.101.103

6F=3+99.101.103

6F=3+1029897

6F=1029900

F =1029900:6

F=171650

=1/2(2/1.3+2/3.5+2/5.7+....+2/2009.2011

=1/2(1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-1/2011

=1/2(1/1-1/2011)

=1/2.2010/2011

=1005/2011

=1/1-1/3+1/3-1/5+1/5-1/7+....+1/2009-2011

=1-1/2011

=2010/1011

cậu có kết bạn với tớ không

Ta có: A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2013.2015}\)

\(\Leftrightarrow2A=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2013.2015}\right)\)

\(\Leftrightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2013}+\dfrac{1}{2013}-\dfrac{1}{2015}\)

\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{2015}=\dfrac{2012}{6045}\)

\(\Leftrightarrow A=\dfrac{1006}{6045}\)

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{2013.2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}+\dfrac{1}{2015}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{2015}\)

2A=\(\dfrac{2014}{2015}\)

 A=\(\dfrac{1007}{2015}\)

                     Khi gặp bài này, bn nên tách 1 phân số ra thành hiệu của 2 phân số.

Giải:

A=1/1.3+1/3.5+1/5.7+...+1/2013.2015

A=1/2.(2/1.3+2/3.5+2/5.7+...+2/2013.2015)

A=1/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/2013-1/2015)

A=1/2.(1/1-1/2015)

A=1/2.2014/2015

A=1007/2015

Chúc bạn học tốt!

Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow\frac{x+2}{2021}=1\)

\(\Leftrightarrow x=2019\)

Vậy \(x=2019\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

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

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

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

\(=\frac{4}{9}\)

Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9

2A=2/1.3+2/3.5+2/5.7+2/7.9

2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2A=1-1/9

2A=8/9

A=4/9

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}=\frac{4}{9}\)

k mik nha , mik k lại !!!!!!!!!!!!!!!!!!!!!!!