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

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

A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)

A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)

A           = \(\dfrac{127}{128}\)

Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B

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

\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)

\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)

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

\(B=1+0-\dfrac{1}{128}\)

\(B=1-\dfrac{1}{128}\)

\(B=\dfrac{128}{128}-\dfrac{1}{128}\)

\(B=\dfrac{127}{128}\)

\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

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

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

Vậy \(A=\frac{255}{512}\)

=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152

=1/2-1/152

=255/512

A=255/512

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(A=\frac{2-1}{4}+\frac{2-1}{8}+\frac{2-1}{16}+\frac{2-1}{32}+\frac{2-1}{64}+\frac{2-1}{128}+\frac{2-1}{256}+\frac{2-1}{512}\)

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

\(A=\frac{1}{2}-\frac{1}{512}\)

\(A=\frac{256}{512}-\frac{1}{512}=\frac{255}{512}\)

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

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

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

= 127/ 128

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

= 1 - 1/128 

= 127/128

k nha bn

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

     2A=1+1/2+1/4+1/8+1/16+1/32+1/64

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

     A=1-1/128=127/128
 

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\)

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

\(A=1-\dfrac{1}{128}=\dfrac{127}{128}\)

1/2 + 1/4 + 1/8 + … + 1/128

= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + … + 1/64 - 1/128

= 1 - 1/128

= 128/128 - 1/128

= 127/128

Chúc bạn học tốt.

😁😁😁

cảm ơn nguyễn phú  tài

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

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

A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)

A \(\times\)( 2-1) = \(\dfrac{255}{128}\)

A = \(\dfrac{255}{128}\)

Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T

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

\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)

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

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

\(T=2+0+0+...-\dfrac{1}{128}\)

\(T=\dfrac{256}{128}-\dfrac{1}{128}\)

\(T=\dfrac{255}{128}\)

anh_hung_lang_la thì làm đi

bọn kia có trả lời câu hỏi không thì bảo

1)=1023/1024

2)=1093/729

c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)

\(=\dfrac{255}{103}\)