Đặt A= 1/3+1/9+1/27+1/81+1/243

A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5

3A=1+1/3+1/3^2+1/3^3+1/3^4

3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5

2A=1-1/3^5

2A=242/243

A=121/243

ta bằng 121/243

duyệt nha

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

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

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

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

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

\(A=\frac{1-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{243}}{2}=\frac{121}{243}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= 2 - 1/243

=485/243

là121/243

NÀY CÁC BẠN ƠIII

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

bạn quy đồng mẫu số tất cả cả các phân số trên lên thành 243 rồi cộng vào là được

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

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

\(C=\frac{81+27+9+3+1}{243}=\frac{121}{243}\)

Vậy C = ...

                                                               Bài giải

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

\(C=\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3\cdot3}\)

\(3C=1+\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3} +\frac{1}{3\cdot3\cdot3\cdot3}\)

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

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

\(C=\frac{242}{243}\text{ : }2=\frac{121}{243}\)