\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(3\times A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)

\(2\times A=1-\frac{1}{729}=\frac{728}{729}\)

\(A=\frac{364}{729}\)

1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy: 
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243

tick đúng nha

=364/243

đà đà na đa

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

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

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

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

\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(3B-B=1-\frac{1}{243}\)

\(2B=\frac{242}{243}\)

\(B=\frac{242}{243}\div2\)

\(B=\frac{121}{243}\)

a.A=1/2+1/4+1/8+1/16+1/32+1/64

 A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)

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

= 1 - 1/8 = 7/8

b.B=1/3+1/9+1/27+1/81+1/243

B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)

= 1 - 1/27 = 26/27

😵

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

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

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

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

        A       =  1 - \(\dfrac{1}{32}\)

        A       =   \(\dfrac{31}{32}\)

      B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)

3 \(\times\)B =  1 + \(\dfrac{1}{3}\) +   \(\dfrac{1}{9}\) +  \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)

3 \(\times\) B - B = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) -(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+...+\(\dfrac{1}{729}\))

2B          = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) - \(\dfrac{1}{3}\) - \(\dfrac{1}{9}\)-\(\dfrac{1}{27}\)-...- \(\dfrac{1}{729}\)

2B          = 1 - \(\dfrac{1}{729}\)

2B         = \(\dfrac{728}{729}\)

   B = \(\dfrac{728}{729}\) : 2

   B  = \(\dfrac{364}{729}\)

bài 1 tính nhanh

mik xin sửa đề câu a thành thế này ~

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

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

 \(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) 

\(A\cdot2-A=\) (  \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) )  - (  \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\) )

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

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

\(b,\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

đặt \(B=\) \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) 

     \(B\cdot3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(B\cdot3-B=\)  ( \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\) ) 

\(B\cdot2=\) \(1-\frac{1}{729}\)

\(B\cdot2=\frac{728}{729}\)

\(B=\frac{728}{729}:2\)

\(B=\frac{364}{729}\) 

\(c,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

ĐẶT \(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

    \(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(C=\frac{1}{1}-\frac{1}{6}\)

\(C=\frac{5}{6}\)

Cảm ơn bạn nhé

\(3A=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

Lấy \(3A-A=1-\dfrac{1}{729}\)

\(\Rightarrow2A=\dfrac{728}{729}\Rightarrow A=\dfrac{364}{729}\)

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

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

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

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

2A=1-1/729

2A=728/729

A=728/729/2

A=364/729

Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

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

\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)

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

\(3A-A=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)

\(2A=3-\frac{1}{3^5}\)

\(A=\frac{3-\frac{1}{3^5}}{2}\)

Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

      \(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)

\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)

\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

              \(S\times2=3-\frac{1}{243}\)

              \(S\times2=\frac{728}{243}\)

                    \(S=\frac{728}{243}\div2\)

                    \(S=\frac{364}{243}\)

Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)

khó vậy nhỉ

a) 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 

b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 
3B - B = 1 - 1/729 
2B = 728/729 
B = 364/729

a) 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 

b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 
3B - B = 1 - 1/729 
2B = 728/729 
B = 364/729 

gọi A là biểu thức trên

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

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

      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