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Ta có :
Đặt \(A=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}+\frac{1}{168}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(2A=\frac{1}{2}-\frac{1}{14}\)
\(2A=\frac{7}{14}-\frac{1}{14}=\frac{3}{7}\)
\(A=\frac{3}{7}:2=\frac{3}{14}\)
=> \(\frac{1}{4}+\frac{3}{14}=\frac{7}{28}+\frac{6}{28}=\frac{13}{28}\)
Ủng hộ mk nha !!! ^_^
![](https://rs.olm.vn/images/avt/0.png?1311)
`1/8+1/24+1/48+1/80+1/120`
`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`
`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`
`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`
`=1/2xx(1/2-1/12)`
`=1/2xx(6/12-1/12)`
`=1/2xx5/12=5/24`
\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)
=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
=\(\dfrac{1}{2}.\dfrac{5}{12}\)
=\(\dfrac{5}{24}\)
Dấu chấm(.)là nhân.
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi A=\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}\)
\(2A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{12}\)
\(2A=1-\frac{1}{12}=\frac{11}{12}\)
\(A=\frac{11}{12}:2=\frac{11}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
8 = 2 \(\times\) 4
24 = 4 \(\times\) 6
48 = 6 \(\times\) 8
80 = 8 \(\times\) 10
Xét dãy số: 2; 4; 6; 8;...; đây là dãy số cách đều với khoảng cách là:
4 - 2 = 2
Số thứ 20 của dãy số trên là: 2 x (20 - 1) + 2 = 40
Vậy Phân số thứ 20 của dãy số đã cho là: \(\dfrac{1}{40\times42}\)
Tổng của 20 phân số đầu tiên của dãy số đã cho là:
A = \(\dfrac{1}{8}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{48}\) + \(\dfrac{1}{80}\) +...+ \(\dfrac{1}{1680}\)
A = \(\dfrac{1}{2\times4}\) + \(\dfrac{1}{4\times6}\) + \(\dfrac{1}{6\times8}\) + \(\dfrac{1}{8\times10}\)+...+ \(\dfrac{1}{40\times42}\)
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{2\times4}\) + \(\dfrac{2}{4\times6}\)+\(\dfrac{2}{6\times8}\)+\(\dfrac{2}{8\times10}\)+...+\(\dfrac{2}{40\times42}\))
A = \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{10}\)+...+ \(\dfrac{1}{40}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{42}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{40}{42}\)
A = \(\dfrac{5}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a) \(\frac{6}{15}+\frac{6}{35}+\frac{6}{63}+\frac{6}{99}+\frac{6}{143}\)
\(=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+\frac{6}{11.13}\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{6}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{6}{2}.\frac{10}{39}\)
\(=\frac{10}{13}\)
b) \(\frac{3}{24}+\frac{3}{48}+\frac{3}{80}+\frac{3}{120}+\frac{3}{168}\)
\(=\frac{3}{4.6}+\frac{3}{6.8}+\frac{3}{8.10}+\frac{3}{10.12}+\frac{3}{12.14}\)
\(=\frac{3}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{4}-\frac{1}{14}\right)\)
\(=\frac{3}{2}.\frac{5}{28}\)
\(=\frac{15}{56}\)
\(a.\frac{6}{3.5}+\frac{6}{5.7}+...+\frac{6}{11.13}\)
\(=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=3.\frac{10}{39}\)
\(=\frac{10}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}\)
\(=\frac{1}{2}(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12})\)
\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{10}-\frac{1}{12})\)
\(=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{12}\right)=\frac{1}{2}\cdot\frac{5}{12}=\frac{5}{24}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+\frac{1}{80}+\frac{1}{120}+\frac{1}{168}\)
\(=\frac{1}{4}.\left(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\left(2-\frac{1}{7}\right)\)
\(=\frac{1}{4}.\frac{13}{7}=\frac{13}{28}\)
Là 1 phần 4 + 1phan 8+ va cu nhu the