=127/128

~ chúc bn hok tốt ~

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

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

\(=\frac{126}{128}=\frac{63}{64}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}=\frac{63}{64}\)

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

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

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

\(=\frac{7}{2}\)

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

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

\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)

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!

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

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

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

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

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

Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

2A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

2A - A = \(1-\frac{1}{64}\)

=> A = \(\frac{63}{64}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

=\(=\frac{63}{64}\)

\(\frac{63}{64}\)

bang 63/64 nha 

dung100*** luon

Nhân 2 lên rồi lấy kết quả vừa nhận trừ đi ban đầu

63/64 nha ban

Cách 1:

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

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

A = 2A - A = \(1-\frac{1}{128}\)

=> A = \(\frac{127}{128}\)

Cách 2:

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

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

= \(1-\frac{1}{128}\)

= \(\frac{127}{128}\)

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

Gạch 1/4 với 1/4 , 1/8 với 1/8 , 1/16 với 1/16 , 1/32 với 1/32 , 1/64 với 1/64

Còn 1/2 - 1/128 = 63/128 

Đúng thì k cho mình 

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

\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+\frac{1}{32}\times2+\frac{1}{64}\times2+\frac{1}{128}\times2\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)

\(A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)

Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}\)

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

= 63/64

Ta đặt biểu thức trên là A

A x 2 =\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

A x 2 - A = 1 +\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}\)

A = \(1-\frac{1}{64}\)

A =\(\frac{63}{64}\)