a)Đặt \(A=\dfrac{6}{1.4}+\dfrac{6}{4.7}+\dfrac{6}{7.10}+...+\dfrac{6}{97.100}\)

\(3a=3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+...+\dfrac{3}{97}-\dfrac{3}{100}\)

\(=3-\dfrac{3}{100}\)

\(=\dfrac{297}{100}\)

b)Đặt \(B=\dfrac{4}{1.3}+\dfrac{16}{3.5}+\dfrac{36}{5.7}+...+\dfrac{9604}{97.99}\)

\(=2b=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)

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

\(2b=2-\dfrac{2}{99}=\dfrac{198}{99}-\dfrac{2}{99}=\dfrac{196}{99}\)

c) Tương tự! Bạn tự làm nhé!

Ta có: \(S=\dfrac{4}{1\cdot3}+\dfrac{16}{3\cdot5}+\dfrac{36}{5\cdot7}+...+\dfrac{2500}{49\cdot51}\)

\(=1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{3\cdot5}+1+\dfrac{1}{5\cdot7}+...+1+\dfrac{1}{49\cdot51}\)

\(=25+\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{49\cdot51}\right)\)

\(=25+\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=25+\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)

\(=25+\dfrac{1}{2}\cdot\dfrac{50}{51}\)

\(=25+\dfrac{25}{51}\)

\(=25\cdot\dfrac{52}{51}=\dfrac{1300}{51}\)

sai gòi

chứ bn làm đi

A bn lướt xuống dưới mà xem cách làm 

nhưng của bn là cho 3 ra ngoài nha

Giải:

A=3/1.3+3/3.5+3/5.7+...+3/49.51

A=3/2.(2/1.3+2/3.5+2/5.7+...+2/49.51)

A=3/2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

A=3/2.(1/1-1/51)

A=3/2.50/51

A=25/17

B=1/3+1/3²+1/3³+...+1/3⁸

3B=1+1/3+1/3²+...+1/3⁷

3B-B=(1+1/3+1/3²+...+1/3⁷)-(1/3+1/3²+1/3³+...+1/3⁸)

2B=1-1/3⁸

   B=1-1/3⁸ /2

Chúc bạn học tốt!

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

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

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

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

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​