\(\frac{1}{2017.2016}\)-\(\frac{1}{2016.2015}\)-\(\frac{1}{2015.2014}\)-...................-\(\frac{1}{3.2}\)-\(\frac{1}{2.1}\)
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)
\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)
\(=\frac{1}{2016}\)
\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)
\(=0+\frac{1}{2016}=\frac{1}{2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{2016}=\dfrac{1}{2016}\)
Bài 3 : Tính :
A = \(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+....+\frac{1}{1.2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{1.2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
Vậy \(A=\frac{2015}{2016}\).
Mình viết ngược lại cho dễ làm xD
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}+\frac{1}{2015\cdot2016}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\frac{1}{1}-\frac{1}{2016}\)
\(A=\frac{2015}{2016}\)
Sai thì bỏ quá :3
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2015}+\frac{1}{2014}-...-\frac{1}{2}+1\)
=\(-\frac{1}{2016}+1=\frac{2015}{2016}\)
Ta có :\(\frac{-1}{2016.2015}-\frac{1}{2015.2014}-\frac{1}{2014.2013}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(-\left(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(-\left(\frac{1}{2016}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2013}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)
= \(-\left(\frac{1}{2016}-1\right)\)
= \(-\left(-\frac{2015}{2016}\right)\)
= \(-\frac{2015}{2016}\)
Mk làm kĩ lắm rồi. ko tích nữa mk cũng chịu bạn luôn @@
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=-\dfrac{1}{1.2}-\dfrac{1}{2.3}-...-\dfrac{1}{2014.2015}-\dfrac{1}{2015.2016}\)
\(\Rightarrow-F=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=1-\dfrac{1}{2016}=\dfrac{2015}{2016}\)\(\Rightarrow F=\dfrac{-2015}{2016}\)
Giải:
\(F=\dfrac{-1}{2016.2015}-\dfrac{1}{2015.2014}-\dfrac{1}{2014.2013}-\dfrac{1}{2013.2012}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(\Leftrightarrow F=-\left(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+\dfrac{1}{2013.2012}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}+\dfrac{1}{2013.2014}+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\left(\dfrac{1}{1}-\dfrac{1}{2016}\right)\)
\(\Leftrightarrow F=-\dfrac{2015}{2016}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
bai nay ban viet nguoc day so lai roi giai nhu binh thuong la duoc