A = \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{101.103}\)

A = \(\dfrac{1}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{101.103}\))

A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{101}\) - \(\dfrac{1}{103}\))

A = \(\dfrac{1}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{103}\))

A = \(\dfrac{1}{2}\). \(\dfrac{102}{103}\)

A = \(\dfrac{51}{103}\)

Em ơi thừa số thứ ba phải là \(\dfrac{1}{5.7}\) mới đúng em nhé.

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)

dãy phân số mà bảo là dãy số à ?

a; C = \(\dfrac{3}{1.3}\) + \(\dfrac{3}{3.5}\) + \(\dfrac{3}{3.7}\) + ... + \(\dfrac{3}{49.51}\)

   C = \(\dfrac{3}{2}\).(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{49.51}\))

  C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{51}\))

 C = \(\dfrac{3}{2}\).(\(\dfrac{1}{1}\) - \(\dfrac{1}{51}\))

 C = \(\dfrac{3}{2}\).\(\dfrac{50}{51}\)

C = \(\dfrac{25}{17}\)

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

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

\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)

$A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}$

$A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}$

$A=\frac{1}{1}-\frac{1}{99}$

$A=\frac{98}{99}$

=

21​−51​+51​−71​+....+951​−981​

=12−198=21​−981​tự làm tiếp nha ( giống câu a)

A=1 - 1/3+1/3 - 1/5+1/5 - 1/6+...+1/99 - 101+1/101 - 1/103

A=1 - 1/103

A=102/103

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)

\(=\frac{1}{2}.\frac{102}{103}\)

\(=\frac{51}{103}\)

A = 1/1.3+1/3.5+...+1/99.101+1/101.103

 2A  = 2.(1/1.3+...+1/101.103)

        = 2.102/103

        => A= 102/103

biểu thức trên = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}< 1\)

vậy A<1

Ta thấy

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

\(=\left(\frac{1}{1}+\frac{1}{101}\right)\)

\(=\frac{102}{101}\)

\(\Rightarrow A>1\)

\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)

\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)

\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)

\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)

\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(M=11\cdot\left(1-\frac{1}{103}\right)\)

\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)

\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)

\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)

\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)

\(\Leftrightarrow Q=3.\frac{63}{268}\)

\(\Leftrightarrow Q=\frac{189}{268}\)

Câu b) bạn làm tương tự nhé :)

\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)

\(\Leftrightarrow Q=\frac{3}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)

\(\Leftrightarrow Q=1\left(\frac{1}{4}-\frac{1}{67}\right)\)

\(\Leftrightarrow Q=\frac{63}{268}\)

\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)

\(\Leftrightarrow M=\frac{22}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

\(\Leftrightarrow M=\frac{22}{2}\cdot\frac{102}{103}\)

\(\Leftrightarrow M=\frac{1122}{103}\)