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Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\) + \(\dfrac{4}{8.11}\) + ... + \(\dfrac{4}{65.68}\)
7A = \(\dfrac{4.3}{2.5}\) + \(\dfrac{4.3}{5.8}\) + \(\dfrac{4.3}{8.11}\) + ... + \(\dfrac{4.3}{65.68}\)
7A = 4 (\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{65.68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + ... + \(\dfrac{1}{65}\) - \(\dfrac{1}{68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
7A = 4 . \(\dfrac{33}{68}\) = \(\dfrac{33}{17}\)
A = \(\dfrac{33}{17}\) : 7
=> A = \(\dfrac{33}{119}\)
Ta có: \(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+...+\dfrac{68-65}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)=\dfrac{4}{3}.\dfrac{33}{68}=\dfrac{11}{17}\)
a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{202}{203}\)
b: \(=-4\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)
\(=-\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)
\(=\dfrac{-4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)
\(=\dfrac{-4}{3}\cdot\dfrac{504}{1009}=-\dfrac{672}{1009}\)
Đặt :
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+.........+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+............+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+........+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\)
\(\Leftrightarrow3A=\dfrac{1}{2}-\dfrac{1}{3n+2}\)
@Akai Haruma em không hiểu tại sao bài kia chị lại tick cho bạn đó ạ,đề nói chứng minh,mak bạn đó đã làm hết đâu:
\(VT=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{3n-1}+\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(VT=\dfrac{1}{6}-\dfrac{1}{9n+6}\)
\(VT=\dfrac{9n+6}{54n+36}-\dfrac{6}{54n+36}\)
\(VT=\dfrac{9n+6-6}{54n+36}=\dfrac{9n}{54n+36}=\dfrac{9n}{9\left(6n+4\right)}=\dfrac{n}{6n+4}=VP\left(đpcm\right)\)
3A=3/2.5+...+3/2018.2021
3A=1/2-1/5+1/5-...+1/2018-1/2021
3A=1/2-1/2021 sau tự tính A
4/9 x 11 + 4/11 x 13 + 4/13 x 15 + ... + 4/97 x 99
= 2(1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99)
= 2(1/9 - 1/99)
= 20/99
Chúc bạn học tốt!
A =\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
A = \(\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-...-\left(\dfrac{1}{65}-\dfrac{1}{65}\right)-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
1/3.(1/2.5+1.5.8+1/8.11+...+1/65.68)
=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/65-1/68)
=1/3(1/2-1/68)
=1/3.33/68
=11/68
nhớ theo dõi mik nha