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 : 

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

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

\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)

\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

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

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

\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)

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

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

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

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

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

\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)

\(B=-\left(1-\frac{1}{1024}\right)\)

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

chịu

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

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

Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)

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

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

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

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

A= \(\frac{1025}{1024}\)

= 2047/2048 nha bạn k mình nha 

1024 phan 2048

A=1/2+1/4+1/8+1/16+...+1/2048

2A=1+1/2+1/4+1/1/8+...+1/1024

2A-A=(1+1/2+...+1/1024)-(1/2+1/4+...+1/2048)

A=1-1/2048

A=2047/2048

ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

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

tách

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

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

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

thay vào B ta có 

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

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

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

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

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

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

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

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

\(\Rightarrow A=\frac{513}{1024}\)

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

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

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

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

BẤM ĐÚNG NHÉ

1023/1024 nhé bạn

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

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

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

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

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

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