A x 2 = 1 + 1/2+1/4+1/8+1/16+1/32 

A x 2 - A = (1 + 1/2+1/4+1/8+1/16+1/32 ) -  ( 1/2+1/4+1/8+1/16+1/32 +1/64)

A            =   1 - 1/64

A =                    63/64          

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

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

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

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

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

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

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

\(B=1-\frac{1}{16}=\frac{15}{16}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{64}\)

#)Giải :

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

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

\(=\frac{1}{2}-\frac{1}{128}\)

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

Câu a:Đáp số 127/128

Câu b:Đáp số 1021/63

a) Đặt A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 
2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256) 
     = 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 
=>A =  2A-A =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256 
=1-1/256 
=255/256

b) = 18 - (19/21 + 8/9) = 18 - 113/63 = 1021/63

a) Đặt A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 
2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256) 
     = 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 
=>A =  2A-A =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256 
=1-1/256 
=255/256

b) sai đề

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

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

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

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

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

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

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

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

A = 1 - 1/256

A = 255/256

A=$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}$

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

A=$1-\frac{1}{256}$

A=$\frac{255}{256}$

1/2+1/4+1/8+1/10+1/16+1/32+1/64+1/128

=1-1/2+1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128+1/128

=1-1/128

=127/128

#Sagittarius#

1.a 

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

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

= 1-1/32=31/32

1b

\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3} +\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)

\(=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}+\frac{1}{4}+\frac{1}{20}+\frac{1}{30}\)

\(=\frac{5}{20}+\frac{5}{30}+\frac{20}{30}+\frac{5}{20}+\frac{1}{20}+\frac{1}{30}\)

\(=\left(\frac{5}{20}+\frac{5}{20}+\frac{1}{20}\right)+\left(\frac{5}{30}+\frac{20}{30}+\frac{1}{30}\right)\)

\(=\frac{11}{20}+\frac{26}{30}\)

\(=\frac{11}{20}+\frac{13}{15}\)

\(=\frac{17}{12}\)

Đó là các cách bạn Võ Thạch Tín đã nêu ra nên mình ko còn câu trả lời nào hết

ta có:A=1/2+1/2^2+1/2^3+...+1/2^6+1/2^7  (1)

 2A=     2.(1/2+1/2^2+1/2^3+...+1/2^6+1/2^7)

      =1+1/2+1/2^2+....+1/2^6+1/2^7  (2)

  lấy (2) trừ (1) vế với vế ta được:
2A-A=(1+1/2+1/2^2+....+1/2^6+1/2^7)-(1/2+1/2^2+...+1/2^6+1/2^7)

     A=1-1/2^7

  VẬY A=1-1/2^7

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

127/128