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.

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

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

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

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

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

\(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{2021\cdot2022\cdot2023}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{2021\cdot2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{2021\cdot2022}-\dfrac{1}{2022\cdot2023}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4090506}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2045252}{4090506}=\dfrac{1022626}{4090506}=\dfrac{511313}{2045253}\)

`1/(1.2.3) + 1/(2.3.4) + ... + 1/(2021 . 2022 .2023)`

`=> 2/(1.2.3) + 2/(2.3.4) + ... + 2/(2021 . 2022. 2023)`

`= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + ... + 1/(2021.2022) - 1/(2022 . 2023)`

`= 1/2 - 1/4090506`

`=4090506/8181012 - 2/8181012`

`= 4090504/8181012`

Giải:

\(\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+...+\dfrac{5}{98.99}+\dfrac{5}{99.100}\) 

\(=5.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\) 

\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\) 

\(=5.\left(1-\dfrac{1}{100}\right)\) 

\(=5.\dfrac{99}{100}\) 

\(=\dfrac{99}{20}\) 

Chúc em học tốt!

Giải:

=5.(11.2+12.3+13.4+...+198.99+199.100)=5.(11.2+12.3+13.4+...+198.99+199.100) 

=5.(1−1100)=5.(1−1100) 

=9920=9920 

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

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

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

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

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

bài này tương tự bài trên

Ta đặt

  \(A=\dfrac{7}{1\times2}+\dfrac{7}{2\times3}+...+\dfrac{7}{99\times100}\)

\(\dfrac{1}{7}\times A=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+....+\dfrac{1}{99\times100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=1-\dfrac{1}{100}\)

\(\dfrac{1}{7}\times A=\dfrac{99}{100}\)

\(A=\dfrac{99}{100}\div\dfrac{1}{7}\)

\(A=\dfrac{693}{100}\)

= 7.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100)

= 7.(1 - 1/100)

= 7 . 99/100

= 693/100

\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(\Rightarrow A=7x\dfrac{99}{100}=6,93\)

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

\(E=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)

\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)

...

E = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)

2E = \(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\)

= \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}\right)\)+\(\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\) +...+ \(\left(\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{99.100}\)= \(\dfrac{1}{2}-\dfrac{1}{9900}\) = \(\dfrac{4949}{9900}\)

a) Ta có

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

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

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

S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)

b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)

A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

A = \(2-\dfrac{1}{99}\)

A = \(\dfrac{197}{99}\)

c) Ta có

B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)

B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

B = \(1-\dfrac{1}{100}\)

B = \(\dfrac{99}{100}\)

d) Ta có

C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)

C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)

C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)

C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))

Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)

D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

D = \(\dfrac{1}{2}-\dfrac{1}{99}\)

D = \(\dfrac{97}{198}\)

=> C = 51 + 100.\(\dfrac{97}{198}\)

C = 51 + \(\dfrac{4850}{99}\)

C = \(\dfrac{9899}{99}\)

Đây là bài làm của mình sai thì nx nha