![](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)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{99.100}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}\)
\(=\frac{99}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=3/1.2+3/2.3+3/3.4+3/4.5+...+3/2021.2022
A=3(1/1.2+1/2.3+1/3.4+1/4.5+...+1/2021.2022)
A=3(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2021-1/2022)
A=3[1/1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/2021-1/2021)-1/2022]
A=3[1/1+0+0+0+...+0-1/2022
A=3(1/1-1/2022)
A=3(2022/2022-1/2022)
A=3.2021/2022
A=2021/674
Bn Tham Khảo:
https://hoc247.net/hoi-dap/toan-6/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-2016-faq188428.html
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{2014}{2015}\)
\(\Leftrightarrow A=\frac{2014}{2015}\div\frac{1}{4}\)
\(\Leftrightarrow A=\frac{8056}{2015}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt S=1.2+2.3+.........+2011.2012
3S=1.2.3+2.3.(4-1)+...........+2011.2012.(2013-2010)
3S=1.2.3+2.3.4-1.2.3+...........+2011.2012.2013-2010.2011.2012
3S=2011.2012.2013
S=2011.2012.2013:3
S=2714954572
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2011.2012
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)
=> 3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012
=> 3A = 2011.2012.2013
=> A = \(\frac{2011.2012.2013}{3}=2714954572\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{7}{1.2}+\dfrac{7}{2.3}+\dfrac{7}{3.4}+...+\dfrac{7}{2011.2012}\)
\(A=7\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2011.2012}\right)\)
\(A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\right)\)
\(A=7\left(1-\dfrac{1}{2012}\right)=7.\dfrac{2011}{2012}=\dfrac{14077}{2012}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
đặt biểu thức là A ta có:
A= 1.2+2.3+...+100.101
3A=1.2.3+2.3.3+...+100.101.3
3A=1.2.3+2.3.(4-1)+...+100.101.(102-99)
3A=1.2.3+2.3.4-1.2.3+...+100.101.102-99.100.101
3A=100.101.102
A=100.101.102/3
A=343400
duyệt nha
A= 1.2+2.3+....+100.101
3A=1.2.3+2.3.3+....+100.101.3
3A=1.2.(3-0)+2.3.(4-1)+.....+100.101.(102-99)
3A=1.2.3-0+2.3.4-1.2.3+..+100.101.102-99.100.101
3A=100.101.102
A=343400