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

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

\(\Rightarrow3A-A=1-\frac{1}{729}\)

\(\Rightarrow2A=\frac{728}{729}\)

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

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

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

\(\Rightarrow2A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)

\(\Rightarrow2A-A=\frac{1}{3^1}-\frac{1}{3^7}\)

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

Mình giúp bạn nè  

Ta có:

\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(\Rightarrow3A-A=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)

\(\Rightarrow2A=3-\frac{1}{2187}=\frac{6561}{2187}-\frac{1}{2187}=\frac{6560}{2187}\)

\(\Rightarrow A=\frac{6560}{2187}:2=\frac{3280}{2187}\)

cảm ơn bạn nhiều lắm

ko có j  

tổng các ps trên là ; \(\frac{364}{729}\)

đặt biểu thức đó là X

ta có :

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

\(\Rightarrow3X-X=1-\frac{1}{729}\)

\(\Rightarrow X=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)

Đặt \(D=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

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

\(\Leftrightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)

\(\Leftrightarrow3D-D=2D=1-\frac{1}{3^6}\)

\(\Leftrightarrow D=\left(1-\frac{1}{3^6}\right)\div2\)

Vậy kết quả của bạn là gì , zZz NCTK zZz ?

Đặt  \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

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

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

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

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

\(\Rightarrow A=\left(\frac{1}{3}-\frac{1}{3^8}\right):\frac{2}{3}\)

Chúc bạn học tốt !!!

P/s: Bài này đáp số ghi vậy nha bạn không tính ra kết quả được

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

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

=\(\frac{3^6}{3^6}+\frac{3^5}{3^6}+\frac{3^4}{3^6}+\frac{3^3}{3^6}+\frac{3^2}{3^6}+\frac{3^1}{3^6}+\frac{3^0}{3^6}\)

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

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

=\(\frac{1093}{729}\)

nha.

tong cua day so tren la 1093/729

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

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

=1093/729

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

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

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

\(=\frac{1093}{729}\)

gọi biểu thức trên là A

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

          \(\frac{1}{3}\)x  A =\(\frac{1}{3}\)+\(\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)   (2)           

lấy (1) - (2)

           \(\frac{2}{3}xA\)=  1 - \(\frac{1}{2187}\)

            \(\frac{2}{3}xA\)= \(\frac{2186}{2187}\)

                  A       =  \(\frac{2186}{2187}:\frac{2}{3}\)

                  A       =      \(\frac{1093}{729}\)

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

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

\(\Rightarrow3A-A=3-\frac{1}{729}\)

\(\Rightarrow2A=\frac{2186}{729}\)

\(\Rightarrow A=\frac{1093}{729}\)

Đặt B = 1/3 + 1/9 + 1/27 + 1/81 +1/243 + 1/729 + 1/2187

B x 3 = 3 x ( 1/3 + 1/9 +.......+ 1/729 + 1/2187)

        =  1 + 1/3 + 1/9 +.........+1/243 +1/729 

Lấy B x 3 - B ta có :

B x 3 - B = 1 + 1/3 +1/9+ .........+1/243 + 1/729 - 1/3 + 1/9 +.........+1/729 +1/2187

B x (3 - 1)= 1 - 1/2187

B x  2     =   2186/2187

B       = 2186/2187 : 2 = 1093/2187

dễ mk nhìn là biết

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

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

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

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

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

a, Gọi biểu thức đó là A

Ta có :

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

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

A x 3 = \(1+A-\frac{1}{729}\)

A x 3 = \(\frac{728}{729}+A\)

A x 2 + A = \(\frac{728}{729}+A\)

A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)

A = \(\frac{728}{729}\div2=\frac{364}{729}\)

Đáp án = \(\frac{364}{729}\)

b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)