Gọi tong trên là A

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)

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

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

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

\(2A=1-\frac{1}{2187}\)

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

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

\(A=\frac{1093}{2187}\)

Vậy tổng A = \(\frac{1093}{2187}\)

\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)

     \(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)

=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)

<=> 2y = 3- 1/2187

=> y = \(\frac{3-\frac{1}{2187}}{2}\)

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

\(\Rightarrow2187A=2187+729+243+81+27+9+3+1\)

\(\Leftrightarrow2187A=3280\)

\(\Leftrightarrow A=\frac{3280}{2187}\)

Chắc chắn 100% luôn

A=1/3+1/9+1/27+...+1/2187
   =1/3+1/3^2+1/3^3+...+1/3^7
-->3A=1+1/3+1/3^2+...+1/3^6
-->3A-A=(1+1/3+1/3^2+...+1/3^6) - (1/3+1/3^2+1/3^3+...+1/3^7)
-->2A=1- 1/3^7
-->A=1093/2187

ko hiểu có thể giải lại dc ko bạn

pn ơi cho mk hỏi các pn học mũ (x²) chua

#)Giải :

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

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

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

\(A=\frac{3037}{2187}\)

         #~Will~be~Pens~#

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

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

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

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

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

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

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

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

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

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

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

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

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

\(\Rightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

\(\Rightarrow3S-S=1-\frac{1}{3^7}\)

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

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

Câu a

\(S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+\frac{7-5}{5x7}+...+\frac{2019-2017}{2017x2019}.\)

\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}=1-\frac{1}{2019}=\frac{2018}{2019}\)

Câu b

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

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

\(2A=3A-A=1-\frac{1}{3^7}\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^7}\)

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