![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}\)
\(=\frac{51}{103}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(Q=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{64}-\frac{1}{67}\)
\(Q=\frac{1}{4}-\frac{1}{67}=\frac{63}{268}\)
\(M=\frac{22}{1.3}+\frac{22}{3.5}+...+\frac{22}{101.103}\)
\(M=\frac{22}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(M=11\cdot\left(1-\frac{1}{103}\right)\)
\(M=11\cdot\frac{102}{103}=\frac{1122}{103}\)
\(Q=\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{64.67}\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3\left(\frac{1}{4}-\frac{1}{67}\right)\)
\(\Leftrightarrow Q=3.\frac{63}{268}\)
\(\Leftrightarrow Q=\frac{189}{268}\)
Câu b) bạn làm tương tự nhé :)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{101.103}\)
=\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{101.103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{101}-\frac{1}{103}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)
=\(\frac{1}{2}.\frac{102}{103}\)
=\(\frac{51}{103}\)
51/103