Ta có : 

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)

\(A=\frac{2^{10}-1}{2^{10}}\)

\(A=\frac{1024-1}{1024}\)

\(A=\frac{1023}{1024}\)

Vậy \(A=\frac{1023}{1024}\)

Chúc bạn học tốt ~ 

Đặt tổng trên là A.

Ta có

A x 2 = 1+ 1/2+1/4+1/8+ 1/16+1/32+ 1/64+ 1/128 + 1/256 + 1/512

Ax2 - A = 1+ 1/2+1/4+1/8 +1/16 + 1/32 +1/64+ 1/128 + 1/256+ 1/512 - ( 1/2 + 1/4 +1/8+1/16+1/32+1/64 + 1/128+ 1/256 + 1/512+ 1/1024)

A = 1+ 1/2 +1/4+1/8+1/16+1/32+1/64+1/128+1/256 + 1/512 - 1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512- 1/1024

A = 1 - 1/ 1024 = 1023/1024

Đáp án là : 1023/1024

kb nha

học tốt

\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)

\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)

#)Giải :

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

\(\Rightarrow2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

\(\Rightarrow2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=\frac{1}{2^{11}}-\frac{1}{2}\)

\(\Rightarrow A=-\frac{1023}{2048}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)

\(=1-\frac{1}{1024}\)

\(=\frac{1023}{1024}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}.\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

<=> \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)

<=> \(2A-A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{256}+\frac{1}{512}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}-\frac{1}{1024}\)

<=> \(A=1-\frac{1}{1024}\)

<=> \(A=\frac{1023}{1024}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)

BẤM ĐÚNG NHÉ

1023/1024 nhé bạn

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+...+\frac{1}{512}\times2+\frac{1}{1024}\times2\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)

\(A=1-\frac{1}{1024}\)

\(A=\frac{1023}{1024}\)

Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)

Lấy (2) trừ (1) theo vế ta có : 

2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)

Xin lỗi bạn Trần thị mai Chi nha mk bấm sai kết quả . Kết quả đúng là : 

\(A=\frac{2^{10}-1}{2^{10}}\)

Ta có: \(\frac{1}{2}=1-\frac{1}{2}\);  \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\); \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\);  ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\); \(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)

Vậy \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

            \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)

            \(=1+1-\frac{1}{1024}\)

            \(=2-\frac{1}{1024}=\frac{2047}{1024}\)

bằng 2047/1024

\(\frac{2047}{1024}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)

\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

\(A=2^{11}-2\)

(1981 x 1982 - 990) : (1980 x 1982 + 992)

=(1980 x 1982+1982 -990) : (1980 x 1982 +992)

=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)

=1

\(A=\left(1-\frac{1}{2^1}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^9}\right)+\left(1-\frac{1}{2^{10}}\right)\)

\(A=\left(1+1+1+...+1+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

                        10 số 1

\(A=10-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)

\(2B-B=1-\frac{1}{2^{10}}=B\)

=> \(A=10-\left(1-\frac{1}{2^{10}}\right)\)

=> \(A=10-1+\frac{1}{2^{10}}\)

=> \(A=9\frac{1}{1024}\)