a) A = 3² . 1/243 . 81² . 1/3²

= 3² . 1/3⁵ . (3⁴)² . 1/3²

= 3² . 1/3⁷ . 3⁸

= 3¹⁰ . 1/3⁷

= 3³

= 27

b) B = (4.2⁵) : (2³ . 1/6)

= (4.32) : (8 . 1/6)

= 128 : 4/3

= 96

c)  C = (-1/3)³.(-1/3)².(-1/3)

= (-1/3)³⁺²⁺¹

= (-1/3)⁶

= 1/729

d) D = (-1/3)⁻¹ - (-6/7)⁰ + (1/2)² : 2

= -3 - 1 + 1/4 : 2

= -4 + 1/8

= -31/8

cảm ơn

\(=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}\)

\(=9\)

a: \(3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^2}\)

\(=\frac{3^2}{3^5}\cdot\frac{3^8}{3^2}=\frac{1}{3^3}\cdot3^6=3^3=27\)

b: \(4^6\cdot256^2\cdot2^4=\left(2^2\right)^6\cdot\left(2^8\right)^2\cdot2^4\)

\(=2^{12}\cdot2^{16}\cdot2^4=2^{12}\cdot2^{20}=2^{32}\)

c: \(4^6\cdot9^5+6^9\cdot120\)

\(=2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5\)

\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5\)

\(=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)

Ta có \(8^4\cdot3^{12}-6^{11}\)

\(=2^{12}\cdot3^{12}-2^{11}\cdot3^{11}\)

\(=2^{11}\cdot3^{11}\left(2\cdot3-1\right)=2^{11}\cdot3^{11}\cdot5\)

Ta có: \(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac45\)

d: \(4^2\cdot25^2+32\cdot125\)

\(=2^4\cdot5^4+2^5\cdot5^3\)

\(=2^4\cdot5^3\left(5+2\right)=2^4\cdot5^3\cdot7\)

\(B=\frac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\)

\(=\frac{2^4\cdot5^3\cdot7}{2^3\cdot5^2}=2\cdot5\cdot7=70\)

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

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

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

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

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

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

812 x 5 + 812 x 7 - 7 - 812 x 2 ???

 toàn câu hỏi sai

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

đặt biểu thức đó là X

ta có :

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

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

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

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

\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)

\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)

\(2G=3-\frac{1}{3^5}\)

\(2G=3-\frac{1}{243}\)

\(2G=\frac{729}{243}-\frac{1}{243}\)

\(G=\frac{728}{243}:2\)

\(G=\frac{364}{243}\)

\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)

\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)

\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)

\(1-\frac{1}{x-1}=\frac{2014}{2015}\)

\(\frac{1}{x-1}=1-\frac{2014}{2015}\)

\(\frac{1}{x-1}=\frac{1}{2015}\)

\(\Rightarrow x-1=2015\)

\(\Rightarrow x=2016\)

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

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

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

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

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

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

nha.

tong cua day so tren la 1093/729

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

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

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

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

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       =      \(\frac{1093}{729}\)