\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

\(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{12}{37}\)

a) \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot13}+...+\frac{3}{647\cdot650}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}=\frac{1}{5}-\frac{1}{650}=\frac{129}{650}\)

b) \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)

sửa 199 -> 200

P/S : Lần sau đừng có đăng từng câu từng câu hỏi trên đây nhá

                                                       Bài giải

a, \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{647\cdot650}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}\)

\(A=\frac{1}{5}-\frac{1}{650}=\frac{13}{650}-\frac{1}{650}=\frac{12}{650}=\frac{6}{325}\)

b, \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}\)

\(B=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)

\(B=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)\)

\(B=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)

Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

\(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)(x phải khác \(-\frac{3}{4};-\frac{7}{4}\)nhé)

\(\Leftrightarrow\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4x+3\right)\left(4x+7\right)}=4.\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4x+3}-\frac{1}{4x+7}=\frac{5}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{4x+7}=\frac{5}{3}\)

\(\Leftrightarrow\frac{4x+7-3}{3\left(4x+7\right)}=\frac{5\left(4x+7\right)}{3\left(4x+7\right)}\)

\(\Rightarrow4x+7-3=20x+35\)(chỗ này dùng dấu suy ra nhé)

\(\Leftrightarrow4x-20x=35-7+3\)

\(\Leftrightarrow-16x=31\)

\(\Leftrightarrow x=-\frac{31}{16}\)

V...

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069

Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)

\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)

\(=2-\frac{1}{5}\)< 2 

Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2

\(K=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)

\(=\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{89}\right)=\dfrac{5}{4}\cdot\dfrac{86}{267}=\dfrac{215}{534}\)

Ta thấy \(\frac{1}{3}-\frac{1}{7}=\frac{7-3}{3.7}=\frac{4}{3.7}\) 

             \(\frac{1}{7}-\frac{1}{11}=\frac{11-7}{7.11}=\frac{4}{7.11}\)

             ..........................

             \(\frac{1}{1023}-\frac{1}{1027}=\frac{1027-1023}{1023.1027}=\frac{4}{1023.1027}\)

=> \(\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{1023.1027}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{1023}-\frac{1}{1027}\)

=>                                                       =\(\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3.1027}\)

Ta có: \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{1023}-\frac{1}{1027}\)

\(=\frac{1}{3}-\frac{1}{1027}=\frac{1024}{3081}\)

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{1023.1027}\)

\(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{1023.1027}\)

\(=\frac{1}{3}-\frac{1}{1027}\)

\(=\frac{1024}{3081}\)

ai k mik mik k lại nha