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

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

A = 217/218 tick đúng nha

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

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

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

\(A=\left(\frac{1}{2}-\frac{1}{256}\right)\times2=1-\frac{1}{128}=\frac{127}{128}\)

B)A*2=(1/2+1/4+....+1/256)*2

=1+1/2+1/4+....+1/128)

A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)

=1-1/256

=255/256

a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)

  \(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)

Lấy \(A-\frac{1}{3}\times A\)theo vế ta có : 

\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)

\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)

\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)

  \(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)

  \(\Rightarrow A=\frac{605}{162}\)

Vậy  \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)

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

=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)

Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có : 

\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)

\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)

\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)

\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)

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

Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)

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

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

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

               \(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\right)\)

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

           \(A=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)

          \(\Rightarrow\frac{63}{256}.x=\frac{1}{512}=\frac{1}{2^9}\)

           \(\Rightarrow\frac{63}{2^8}.x=\frac{1}{2^9}\)

            \(\Rightarrow x=\frac{1}{2^9}:\frac{63}{2^8}=\frac{1}{2^9}.\frac{2^8}{63}=\frac{1}{2.63}=\frac{1}{126}\)

           Ủng hộ mk nha !!! ^_^

Bài 1:

Bài 1: 1/3+1/9+1/27+1/81+1/243+1/729

Đặt:
A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Nhân A với 3 ta có:
\(Ax3=3+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow Ax3-S=3-\frac{1}{243}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2\)
\(\Rightarrow A=\frac{1093}{729}\)

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

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

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

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

Lấy A x 2 - A ta có :

A x 2 - A = ​1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256

A x ( 2 - 1 ) = 1 - 1/ 256

A               =           255/256

b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729

A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)

        = 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243

Lấy A x 3 - A ta có : 

A x 3 - A = 1 + 1/3 + 1/9 +.....  + 1/243 -  1/3 + 1/9 +........+ 1/243 + 1/29

A x ( 3 - 1 ) = 1 - 1/29

A x2          =    28/29

A = 28/29 : 2 ( tự tính

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)

\(\Rightarrow2A-A=1-\frac{1}{256}\)

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

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

A=127/128

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

suy ra: 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

hok tốt

đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)

=> A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^8}\)

=> 2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^7}\)

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

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

#)Giải :

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

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

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

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

Lời giải 

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

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

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