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\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\div\frac{4+\frac{4}{7}+\frac{4}{9}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{343}}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\div\frac{4\left(1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}\right)}{1+\frac{1}{7}+\frac{1}{9}+\frac{1}{3}}\)

\(=\frac{1}{2}\div4=\frac{1}{8}\)

câu này đơn giản lắm

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4+\frac{4}{7}+\frac{4}{49}+\frac{4}{343}}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\)

\(B=\frac{1}{2}:4=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}\)

= \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}+\frac{1}{343}}\right):\frac{91}{80}\)

= \(\frac{1}{2}:4:\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

                                                Bài giải

      \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}\text{ : }\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{919191}{808080}\)

\(=\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}\text{ : }\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\text{ : }\frac{91}{80}\)

\(=\left(\frac{1}{2}\text{ : }\frac{4}{1}\right)\text{ : }\frac{91}{80}=\frac{1}{8}\text{ : }\frac{91}{80}=\frac{10}{91}\)

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

      \(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)

Từ đề bài ta có

\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)

\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)

\(D=182\times\frac{1}{8}\times\frac{80}{91}\)

\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)

cho tui nha

Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)

\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)

Vậy D=20
 

BẰNG 20 ĐÓ

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

đặt biểu thức đó là X

ta có :

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

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

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

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

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

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

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

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

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

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

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

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

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

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