tham khảo

Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729) 
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 
A x 3 - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729) 
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729 
= 1 - 1/729 
A x 2 = 728/729 
A = 364/729

\(\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{2}{729}+\dfrac{1}{729}=\dfrac{243+81+27+9+2+1}{729}=\dfrac{363}{729}=\dfrac{121}{243}\)

\(3A=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)

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

Lấy \(3A-A=1-\dfrac{1}{729}\)

\(\Rightarrow2A=\dfrac{728}{729}\Rightarrow A=\dfrac{364}{729}\)

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

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

3A-A=(1+1/3+1/9+1/27+1/81+1/243)-(1/3+1/9+1/27+1/81+1/243+1/729)

3A-A=1-1/3+1/3-1/9+1/9-1/27+1/27-1/81+1/81-1/243+1/243-1/729)

2A=1-1/729

2A=728/729

A=728/729/2

A=364/729

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

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

=>3S-S=\(\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)

=>s=1-1/729 = 728/729

1/3+1/9+1/27+1/81+1/243+1/729=(1/3+1/9+1/81)+(1/27+1/243+1/729)=37/81+37/729=333/729+37/729=370/729

= 370/729

Điểm âm của mik chỉ còn 109 nữa thôi, các bn giúp mik nha

ai giúp mik, mik sẽ giúp lại

cảm ơn trc~~~

364/729 nha bạn

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

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

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

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

\(A=\frac{728}{729}:2=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)

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

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

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

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

\(A=\frac{728}{729}:2=\frac{728}{729}.\frac{1}{2}=\frac{364}{729}\)

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

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

=1093/729

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

=1093/729

làm từng bước đi nhé bạn.kết quả mình cũng ra rồi nhưng mình muốn tham khảo cách làm của các bạn

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

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

= 2 - 1/243

=485/243

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

=1093/729

A=1+1/3+1/3^2+......1/3^6

3A= 3 +1 + 1/3+......=1/3^5

3A-A= 3-1/3^6

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

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

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

\(A\)\(x3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(A\)\(x3=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+3\)

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

\(A\)\(x2=3-\frac{1}{729}=\frac{2186}{729}\)

\(A=\frac{2186}{729}:2=\frac{1093}{729}\)

1093/729

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

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

\(3\times A-A=3-\frac{1}{729}=\frac{2186}{729}\)

\(2\times A=\frac{2186}{729}=>A=\frac{1093}{729}\)