Dấu "." là dấu nhân bạn nhé.

Ta có:

\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)

\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)

\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)

\(\Rightarrow2A=3-\dfrac{1}{729.3}\)

\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)

Lời giải:

Đặt $A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}$
$3\times A=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}$

$3\times A-A=3-\frac{1}{2187}$

$2\times A=3-\frac{1}{2187}=\frac{6560}{2187}$

$A=\frac{6560}{2187}:2=\frac{3280}{2187}$

121/81

\(=\frac{121}{81}\)

(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2

Dãy số trên có số số hạng là: (25 - 1): 2 + 1  = 13

Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))

A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)

A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)

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

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

B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)

2B = \(\dfrac{242}{243}\)

B = \(\dfrac{242}{243}\): 2

B = \(\dfrac{121}{243}\)

11a + B = 11a + \(\dfrac{121}{243}\) (2)

Từ (1) và(2) ta có:

a\(\times\)13  + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)

a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\) 

\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)

a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)

a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)

a \(\times\) 2 = \(\dfrac{109}{6075}\)

a = \(\dfrac{109}{6075}\): 2

a = \(\dfrac{109}{12150}\)

Câu trả lời hay nhất: 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
A x 3 ‐ A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 ‐ ﴾1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729﴿
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 ‐ 1/3 ‐ 1/9 ‐ 1/27 ‐ 1/81 ‐ 1/243 ‐ 1/729
= 1 ‐ 1/729
A x 2 = 728/729
A = 364/729

NHỚ TK MK NHA

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

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

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

\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)

\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)

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

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

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

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049

3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )

3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 

3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 

- ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )

= 1 - 1/59049 

2 x A = 59048/59049

A = 59048/59049 : 2

A = 29524/59049

A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049

3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )

3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 

3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 - ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )

= 1 - 1/59049 

2 x A = 59048/59049

A = 59048/59049 : 2

A = 29524/59049

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

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

\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)

\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)

Phải là : A=1/3+1/9+1/27+1/81+1/243 ta có: 3A=1+1/3+1/9+1/27+1/81                     3A-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)=1-1/243       2A=242/243                                             A=242/243:2=121/243

345(789)432):--:  ...