A = 1×3+3×5+5×7+...+ 97×99+99×101

 6A= 1×3×6+3×5×6+5×7×6+...+97×99×6+99×101×6

6A= 1×3×(5+1)+3×5×(7-1)+5×7×(9-3)+...+97×99×(101-95)+99×101×(103-97)

6A = 1×3×5-1×3+3×5×7-1×3×5+5×7×9-3×5×7+7×9×11-5×7×9+,,,+97×99×101-95×97×99+99×101×103-97×99×101

6A= 1×3+99×101×103

6A= 1029900

A= 171650

171650

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

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

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

\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)

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

S =1/1.3-1/2.4+1/3.5-1/4.6+1/5.7 - 1/6.8+1/7.9-1/8.10

\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{7.9}\right)+\frac{1}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{8.10}\right)\)

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

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

\(=\frac{8}{9}+\frac{2}{5}\)

\(=\frac{58}{45}\)

viết đề hẳn hoi đi đề thì xấu còn bày đặt làm càn

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

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

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

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

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

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

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

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

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

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

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

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

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

\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

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

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

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

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

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

=> \(S=\frac{4}{9}+\frac{1}{5}\)

=> \(S=\frac{29}{45}\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)

\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)

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

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

\(2A=\frac{9}{10}\)

\(A=\frac{9}{10}:2=\frac{9}{20}\)

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

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

( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)

=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

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

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

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 \(A< 1\)