gọi biểu thức là A

ta có :

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...\frac{1}{19.21}\)

=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}...\frac{2}{19.21}\)

2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{21}\)

2A = 1 - \(\frac{1}{21}\)

2A = \(\frac{20}{21}\)

A = \(\frac{20}{21}:2=\frac{10}{21}\)

Ta có:\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)=\frac{1}{2}\left(1-\frac{1}{21}\right)=\frac{1}{2}.\frac{20}{21}=\frac{10}{21}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)\(+...+\frac{1}{19.21}\)

=\(\frac{2}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{19.21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{19}-\frac{1}{21}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{21}\right)\)

=\(\frac{1}{2}.\frac{20}{21}\)

=\(\frac{20}{42}=\frac{10}{21}\)

tớ làm câu b thôi, câu a nhân 1/2 lên là đc 

\(A=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}\right)\right]\)

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

\(A=\frac{1}{2}.\left(1-\frac{1}{2n+1}\right)=\frac{1}{2}-\frac{1}{2.\left(2n+1\right)}< \frac{1}{2}\)

p/s: lưu ý không có dấu "=" đâu nhé vì \(\frac{1}{2.\left(2n+1\right)}>0\left(n\text{ thuộc }N\right)\)

\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(A=\frac{100}{101}:2=\frac{50}{101}\)

\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)

\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)

\(x.\frac{1}{3}=\frac{50}{101}\)

$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$

\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)

\(\frac{1}{3}xx=\frac{50}{101}\)

\(x.x=\frac{150}{101}\)

còn lại tự tính

\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)

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

\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)

\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)

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

\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

\(S=\frac{4}{9}-\frac{1}{5}\)

\(S=\frac{11}{45}\)

=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)

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

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

=\(\frac{29}{45}\)

29/45 bạn nhé

=>2B=2/1.3 +2/3.5 +2/5.7+...+2/2003.2005

=>2B=1-1/3+1/3-1/5+1/5-1/7+...+1/2003-1/2005

=>2B=-1/2005

=>B=-1/2005:2=-1/4010

Vậy B= -1/4010

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

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

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

\(=\frac{1}{2}.\frac{2018}{2019}\)

\(=\frac{2018}{4038}\)

\(\Rightarrow\frac{2018}{4038}< \frac{1}{2}\)( lấy máy tính ) 

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

\(\Rightarrow M=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\)

\(\Rightarrow M=1-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2019}{2019}-\frac{1}{2019}\)

\(\Rightarrow M=\frac{2018}{2019}\)

Có \(\frac{2018}{2019}=\frac{2018.2}{2019.2}=\frac{4036}{4038}\)

\(\frac{1}{2}=\frac{1.2019}{2.2019}=\frac{2019}{4038}\)

Mà \(\frac{4036}{4038}< \frac{2019}{4038}\Rightarrow M< \frac{1}{2}\)

Vậy M < \(\frac{1}{2}\)

theo công thức, ta tính đc: 

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

=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )

1-1/3=2/3 chứ ko phải 1-1/3=1/3 đâu nha bạn