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

đặt biểu thức đó là X

ta có :

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

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

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

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

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

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

=\(\frac{3^6}{3^6}+\frac{3^5}{3^6}+\frac{3^4}{3^6}+\frac{3^3}{3^6}+\frac{3^2}{3^6}+\frac{3^1}{3^6}+\frac{3^0}{3^6}\)

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

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

=\(\frac{1093}{729}\)

nha.

tong cua day so tren la 1093/729

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

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

=1093/729

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

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

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

\(=\frac{1093}{729}\)

gọi biểu thức trên là A

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

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

lấy (1) - (2)

           \(\frac{2}{3}xA\)=  1 - \(\frac{1}{2187}\)

            \(\frac{2}{3}xA\)= \(\frac{2186}{2187}\)

                  A       =  \(\frac{2186}{2187}:\frac{2}{3}\)

                  A       =      \(\frac{1093}{729}\)

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

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

\(\Rightarrow3A-A=3-\frac{1}{729}\)

\(\Rightarrow2A=\frac{2186}{729}\)

\(\Rightarrow A=\frac{1093}{729}\)

dễ mk nhìn là biết

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

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

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

a, Gọi biểu thức đó là A

Ta có :

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

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

A x 3 = \(1+A-\frac{1}{729}\)

A x 3 = \(\frac{728}{729}+A\)

A x 2 + A = \(\frac{728}{729}+A\)

A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)

A = \(\frac{728}{729}\div2=\frac{364}{729}\)

Đáp án = \(\frac{364}{729}\)

b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)

\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)

\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)

\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)

⇒B =
2
1.2
1 −
38.39
1
=
741
( ) 18