Sửa đề:1/1*3+2/3*7+3/7*13+4/13*21+5/21*31

=1/2(2/1*3+4/3*7+6/7*13+8/13*21+10/21*31)

=1/2(1-1/3+1/3-1/7+...+1/21-1/31)

=1/2*30/31=15/31

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

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

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

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

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

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

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

=

21​−51​+51​−71​+....+951​−981​

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

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

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

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

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

\(\Rightarrow A>1\)

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

=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{101.103}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)

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

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

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

51/103