Theo đề bài ta có :

\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)

\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(\Leftrightarrow B=1-\frac{1}{256}\)

\(\Leftrightarrow B=\frac{255}{256}\)

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)

\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)

\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(\Rightarrow B=1-\frac{1}{2^8}\)

p/s: bài lớp 6!

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(B=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)

\(\Rightarrow2B-B=1-\frac{1}{2^8}\)

\(B=1-\frac{1}{2^8}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)

=\(1-\frac{1}{256}\)

=\(\frac{255}{256}\)

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

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

= 255/256

\(=1-\frac{1}{256}=\frac{255}{256}\)

Nhận xét :

1/2 = 1 - 1/2   ;   1/4 = 1/2 - 1/4   ;   1/8 = 1/4 - 1/8   ;   .....   ;   1/256 = 1/128 - 1/256

=> A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/128 - 1/256

=> A = 1 - 1/256 = 255/256

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{156}\)
Ta có:
\(\frac{1}{2}=1-\frac{1}{2}\)  \(;\)   \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(;\)   \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)\(;...;\)   \(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{256}{256}-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Vậy \(A=\frac{255}{256}\)
◘_◘ Đúng 100%

quy đồngcác phân số lấy mẫu số là 512 .ta có tử số là 

256 +128 + 64 +32 +16 +8 +4 +2 +1 =495

A =\(\frac{495}{512}\)

cho hỏi làm thế nào để nó ra phân số như thế kia zạ

đáp án là 511/512

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

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

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

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

BẤM ĐÚNG NHÉ

1023/1024 nhé bạn

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

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

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

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

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

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

 A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512 
A = 1 - 1/512 
A = 511/512 
Chúc bạn học giỏi nha!

A= 1/1022 nha bạn

 A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512 
A = 1 - 1/512 
A = 511/512 

\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)

\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)

\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(A=1-\frac{1}{256}=\frac{255}{256}\)

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

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

\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)

\(A=1-\frac{1}{2^8}\)

\(A=\frac{2^8-1}{2^8}\)

\(A=\frac{255}{256}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)

\(A=1-\frac{1}{256}\)

\(A=\frac{255}{256}\)

Dễ lắm bạn à : 

Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}+\frac{1}{256}\)

 \(\Rightarrow2A=2\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}+\frac{1}{128}\)

\(\Leftrightarrow2A-A=2+1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}+\frac{1}{128}-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Leftrightarrow A=2-\frac{1}{256}=\frac{511}{256}\)

đặt A= 1+1/2+1/4+1/8+...+1/128+1/256

2A=2+1+1/2+1/4+...+1/64+1/128

2A-A=A=2-1/256=511/256

ban giai ro ra duoc ko