\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2021.2023}\right)\)

\(=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{4088484}{2021.2023}\)

\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2022.2022}{2021.2023}\)

\(=\dfrac{2.2022}{1.2023}\)

\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)

\(A=\dfrac{4}{1\cdot2}+\dfrac{4}{2\cdot3}+\dfrac{4}{3\cdot4}+...+\dfrac{4}{2022\cdot2023}\\ A=4\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\\ A=4\cdot\dfrac{2022}{2023}\\ A=\dfrac{8088}{2023}\)

giúp mình nhé! ngày mai mình kiểm tra đề cương r.

`A = 4/(1.2) + 4/(2.3) + ....+ 4/(2022.2023)`

`= 4 . (1/(1.2) + 1/(2.3) + ....+ 1/(2022.2023))`

`= 4 . (1 - 1/2 + 1/2 - 1/3 +.....+ 1/2022 - 1/2023)`

`= 4 . (1 - 1/2023)`

`= 4 . 2022/2023`

`= 8088/2023`

\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)

\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)

\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)

=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)

=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)

A=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2014\cdot2015\cdot2016}=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2014\cdot2015}-\dfrac{1}{2015\cdot2016}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2015}\cdot\dfrac{1}{2016}\right)=\dfrac{1}{4}-\dfrac{1}{2\cdot2015\cdot2016}< \dfrac{1}{4}\)

Vậy A<\(\dfrac{1}{4}\)

---bé hơn hoặc bằng tức là chỉ cần xảy ra 1 khả năng cũng thõa mãn nhé---

mình

bài này tương tự bài trên

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)