TN
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TD
1
14 tháng 3 2022
\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
EK
4
25 tháng 5 2021
`A=4/(1.2)+4/(2.3)+4/(3.4)+......+4/(2014.2015)`
`=4(1/(1.2)+1/(2.3)+1/(3.4)+......+1/(2014.2015))`
`=4(1-1/2+1/2-1/3+1/3-1/4+....+1/2014-1/2015)`
`=4(1-1/2015)`
`=4. 2014/2015`
`=8056/2015`
2
22 tháng 3 2021
\(A=\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
\(A=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{1}{1}-\frac{1}{2015}\right)\)
\(A=4\left(\frac{2015-1}{2015}\right)\)
\(A=4.\frac{2014}{2015}\)
... BẠN TỰ LÀM NỐT NHÉ!
4 tháng 3 2022
\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)
4 tháng 3 2022
\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2014.2015}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\\ =4\left(1-\dfrac{1}{2015}\right)\\ =4.\dfrac{2014}{2015}\\ =\dfrac{8056}{2015}\)
DH
24 tháng 2 2018
Mình k ghi lại đề nhé!~~
\(A=\dfrac{4}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2015}\right)\)
\(A=\dfrac{4}{1}.\left(\dfrac{2015-1}{2015}\right)\)
\(A=\dfrac{4}{1}.\dfrac{2014}{2015}\)
\(A=3,998014888\)
\(A\approx4\)
T
10 tháng 3 2018
Ta có: \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(=1\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2014.2015}\right)\)
\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1\left(1-\frac{1}{2015}\right)\)
\(=1\left(\frac{2015}{2015}-\frac{1}{2015}\right)-1\left(\frac{2014}{2015}\right)=\frac{2014}{2015}\)
Vậy.....
10 tháng 3 2018
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+....+\frac{4}{2014.2015}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)
\(A=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)
30 tháng 8 2016
a=1.2+2.3+3.4+4.+....+200.201
3A = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 200.201.(202 - 199)
3A = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + .... + 200.201.202
3A = 200.201 . 202
A = 2706800
30 tháng 8 2016
\(A=1.2+2.3+3.4+...+200.201\)
\(\frac{1}{A}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{200.201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{200}-\frac{1}{201}\)
\(\frac{1}{A}=\frac{1}{1}-\frac{1}{201}=\frac{200}{201}\)
\(A=1:\frac{200}{201}=\frac{1.201}{200}=\frac{201}{200}\)
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2015}\)
\(\Leftrightarrow\frac{1}{4}A=\frac{2014}{2015}\)
\(\Leftrightarrow A=\frac{2014}{2015}\div\frac{1}{4}\)
\(\Leftrightarrow A=\frac{8056}{2015}\)