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11 tháng 8 2015

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

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

2A = 3A - A = \(1-\frac{1}{3^8}\)

=> A = \(\frac{1-\frac{1}{3^8}}{2}\)

26 tháng 4 2017

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

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^8}\)

\(\Rightarrow A=\left(1-\frac{1}{3^8}\right)\div2\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^8\times2}\)

26 tháng 4 2017

A=1/3+1/3^2+1/3^3+...+1/3^8

3A=1+1/3+1/3^2+...+1/3^7

3A-A=1-1/3+1/3-1/3^2+1/3^2-1/3^3+...+1/3^7-1/3^8

2A=1-1/3^8

2A=6560/6561

Suy ra A=3280/6561

nho k cho minh voi nhe

11 tháng 8 2015

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

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

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

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

=>A=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)

12 tháng 8 2015

 

 A=1/3+1/32+1/33+...+1/38

=>3A=1+1/3+1/32+...+1/37

=>3A-A=(1+1/3+1/32+...+1/37)-(1/3+1/32+1/33+...+1/38)

=>2A=1+1/3+1/32+...+1/37-1/3-1/32-1/33-...-1/38

=1-1/38

=\(\frac{6550}{6561}\)

=>A=\(\frac{6560}{6561}:2=\frac{6560}{6561}.\frac{1}{2}=\frac{3280}{6561}\)

3 tháng 3 2019

A= 1/3 + 1/3^2 + ... + 1/3^8

3A= 3. (1/3+ 1/3^2+ ... + 1/3^8)

3A=1+ 1/3 + 1/3^2+ ... +1/3^7

=> 3A - A= (1 + 1/3 + 1/3^2 + ... + 1/3^7) - (1/3 + 1/3^2+ ... + 1/3^8)

=> 2A= 1 - 1/ 3^8

2A= 6560/6561

A= 6560/6561 : 2

A= 3280/6561

4 tháng 5 2015

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

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

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

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

10 tháng 6 2016

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

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

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

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

8 tháng 5 2018

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

\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^8}\)

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

\(\Rightarrow A=\frac{3280}{6561}\)

Vậy \(A=\frac{3280}{6561}\)

8 tháng 5 2018

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

\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^7}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^8}\)

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

Chúc bạn học tốt !!! 

a: =16*20=320

b: =2/3+1/3*1/6:2/3

=2/3+1/18*3/2

=2/3+3/36

=2/3+1/12

=9/12=3/4