HM
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
8 tháng 5 2018
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
8 tháng 5 2018
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
BN
6 tháng 5 2017
\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2016}}\)
\(2C-C=2+1+....+\frac{1}{2^{2016}}-\left(1+\frac{1}{2}+....+\frac{1}{2^{2017}}\right)\)
\(C\left(2-1\right)=2+1+....+\frac{1}{2^{2016}}-1-\frac{1}{2}-...-\frac{1}{2^{2017}}\)
\(C=2-\frac{1}{2^{2017}}=\frac{2^{2018}}{2^{2017}}-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
ok men nha dug 100%
Co cung ko cai dc
6 tháng 5 2017
\(C=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(C=2-\frac{1}{2^{2016}}\)
WH
8 tháng 4 2018
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(B=1-\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
T
2
11 tháng 5 2019
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)
\(A=2-\frac{1}{2^{2012}}\)
13 tháng 4 2017
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
Nên \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Suy ra \(2A-A=2-\frac{1}{2^{2012}}\)hay \(A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
13 tháng 4 2017
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)
=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
=>\(A=2-\frac{1}{2^{2012}}\)
Cô mình chữa bài này rồi nên bạn cứ yên tâm
TN
17 tháng 5 2016
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+...+\frac{2}{2^{2011}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
17 tháng 5 2016
Ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
=> \(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(2A=2+1+...+\frac{2}{2^{2011}}\)
=> \(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2012}\)
HL
1
Ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
=>\(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)
\(A=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+...+\left(\frac{1}{2^{2016}}-\frac{1}{2^{2016}}\right)-\frac{1}{2^{2017}}\)
\(A=1-\frac{1}{2^{2017}}\)
Vậy: \(A=1-\frac{1}{2^{2017}}\)