Lời giải:

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

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

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

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

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

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

=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)

=> \(A=\frac{3^8-1}{2.3^7}\)

\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729\cdot3}\)

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

\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

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

\(2A=3-\frac{1}{3^7}\)

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

Gia trị biểu thức là : (3-1/3^7) / 2

Tk mk nha

Đáp án A

Đáp án : A

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

Dấu "." là dấu nhân bạn nhé.

Ta có:

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

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

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

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

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

Đáp án C