\(\frac{3^{-m}}{81}=27\)

\(=>3^{-m}=27\cdot81\)

\(3^{-m}=2187\)

Vì nếu \(k^{-m}\) thì => k = \(\frac{1}{k^m}\)

mà 2187 \(\in N\)

=> Không tìm được m thỏa mãn yêu cầu đề bài.

\(\frac{3^{-m}}{81}=27\Rightarrow3^{-m}=27.81=2187=3^7\)

\(\Rightarrow-m=7\)

\(\Rightarrow m=-7\)

 Đặt   \(A=\frac{1}{3}+\frac{1}{9}+.......+\frac{1}{59049}\)

  \(3A=3.\left(\frac{1}{3}+\frac{1}{9}+......+\frac{1}{59049}\right)\)

\(3A=1+\frac{1}{3}+........+\frac{1}{19683}\)

\(3A-A=\left(1+\frac{1}{3}+......+\frac{1}{19683}\right)-\left(\frac{1}{3}+\frac{1}{9}+........+\frac{1}{59049}\right)\)

\(2A=1-\frac{1}{59049}\)

\(2A=\frac{59048}{59049}\)

\(A=\frac{59048}{59049}:2\)

\(A=\frac{59048}{118098}\)

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

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

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

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

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

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

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

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

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

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

3^12.(3^4)^11/(3^3)^10.(3^2)^15=3^12.3^44/3^30.3^30=3^56/3^30

=3^17.9^22/3^30.9^15=9^7/3^13=3^14/3^13=3^1=3

giùm nhé bạn

A=$\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+......+\frac{1}{59049}$

3A=$\frac{1}+frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{19683}$

3A-A=2A=1-1/59049=59048/59049

A=59048/118098

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

đặt biểu thức đó là X

ta có :

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

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

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