lấy MS chung là 2187, ta có:

        729 + 243 + 81 + 9 + 3 + 1                         

       ________________________  =      1066/2187

                      2187

1066/2187.

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

T = 5.5 + 6.6 + .... + 30.30 

T = 5.(6 - 1) + 6.(7-1) + ... + 30.(31 - 1) 

T = 5.6 - 5 + 6.7 - 6 + ... + 30.31 - 30 

T = (5.6 + 6.7 + ... + 30.31) - (5 + 6 + ... + 30) 

Đặt A = 5.6+ 6.7 + ... + 30.31  

     B = 5 + 6 + ... + 30 

Ta có : 

3A = 5.6.3 + 6.7.3 + ... + 30.31 . 3 

3A = 5.6.(7-4) + 6.7.(8-5) + ... + 30.31.(32-29) 

3A = 5.6.7 - 4.5.6 + 6.7.8 - 5.6.7 + ... + 30.31.32 - 29.30.31

3A = (5.6.7 + 6.7.8 + ... + 30.31.32) - (4.5.6 + 5.6.7 + ... + 29.30.31) 

3A  = 30.31.32 - 4.5.6 

3A = 29640

A = 29640 : 3 

A = 9880

SSH của B là : (30 - 5) : 1 + 1  = 26 (số hạng) 

Tổng B là : (30 + 5) . 26  : 2  =455 

=> T = A - B = 9880 - 455 = 9425

c, S = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561

S = (3 + 2187) + (9 + 6561) + (27 + 243) + (81 + 729) +  1 

S = 2190 + 6570 + 270 + 810 + 1 

S = (2190 + 810) + 6570 + 270 + 1 

S = 3000 + 6570 + 270 + 1 

S = 9570 + 270 + 1 

S = 9840  + 1 

S = 9841 

Vậy S = 9841 

a, B=68×75+34×50

   B=68×75+68×25

   B=68×(75+25)

   B=68×100

   B=6800

-597871

nhầm nha

viết sai rồi 

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

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

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

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

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

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

Ta có: 

\(S=3.2^0-3^1+3.2^1-3^2+3.2^2+3.2^3-3^3+3.2^4-3^4+...-3^7+3.2^{10}+3.2^{11}-3^8+3.2^{12}\)

\(=3.\left(2^0+2^1+2^2+2^3+2^4+...+2^{10}+2^{11}+2^{12}\right)-\left(3^1+3^2+3^3+...+3^7+3^8\right)\)

Đặt: \(A=2^0+2^1+2^2+...+2^{11}+2^{12}\)

=> \(2.A=2^1+2^2+2^3+...+2^{12}+2^{13}\)

=> \(2.A-A=2^{13}-2^0\)

\(\Rightarrow A=2^{13}-1=8191\)

Đặt: \(B=3^1+3^2+3^3+...+3^8\)

 \(\Rightarrow3.B=3^2+3^3+3^4+...+3^9\)

=> \(3B-B=3^9-3^1=19680\)

=> \(2B=19680\Rightarrow B=9840\)

=> S=3.A-B=3.8191-9840=14733

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}\)

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

\(3A-A=\left[1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right]-\left[\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right]\)

\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)

\(A=\frac{6560}{6561}:2\)

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

Vậy : ...