a) \(\frac{790^4}{79^4}=\frac{79^4.10^4}{79^4}=10^4=10000\)

b) \(\frac{3^2}{0,375^2}=\frac{0,375^2.8^2}{0,375^2}=8^2=64\)

c) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=3^2.3^{-5}.3^8.3^{-3}=3^2=9\)

d) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=2^7:\left(2^3.2^{-4}\right)=2^7:2^{-1}=2^7:\frac{1}{2}=2^8\)

\(b,\)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

\(\left(-3\right).\frac{1}{243}.81^2.3^{\left(-3\right)}\)

\(=\left(-3\right).\frac{1}{3^3}.\frac{1}{3^5}.3^8\)

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

\(=-3\)

1/243

1/243

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

\(=3^2\cdot\frac{1}{3^5}\cdot3^4\cdot\frac{1}{3^3}\)

\(=\left(3^2\cdot3^4\right)\cdot\left(\frac{1}{3^5}\cdot\frac{1}{3^3}\right)\)

\(=3^6\cdot\frac{1}{3^8}\)

\(=\frac{3^6}{3^8}\)

\(=\frac{1}{3^2}=\frac{1}{9}\)

\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)

= \(9.\frac{1}{243}.6561.\frac{1}{27}\)

= \(9\)

b ) \(\left(4,2\right)^5:\left(2^3.\frac{1}{16}\right)\)

= \(\left(\frac{21}{5}\right)^5:\left(8.\frac{1}{16}\right)\)

= \(130691232:\frac{1}{2}\)

= \(130691232\times2\)

= 261382464

Chúc bạn học tốt  !!!

= 

1.a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=\frac{3^2}{3^5}.\frac{\left(3^4\right)^2}{3^3}\)\(=\frac{3^2.3^8}{3^5.3^3}=\frac{3^{10}}{3^8}=3^2=9\)\(\)

b,  \(\left(4.2\right)^5:\left(2^3.\frac{1}{16}\right)=8^5:\left(8.\frac{1}{16}\right)\)\(=8^5:\frac{1}{2}=\left(2^3\right)^5.2=2^{15}.2=2^{16}\)

chúc bn học tốt!

Đặt A= 1/3+1/9+1/27+1/81+1/243

A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5

3A=1+1/3+1/3^2+1/3^3+1/3^4

3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5

2A=1-1/3^5

2A=242/243

A=121/243

ta bằng 121/243

duyệt nha

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

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

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

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

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

\(A=\frac{1-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{243}}{2}=\frac{121}{243}\)

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

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

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

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

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

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

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

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

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

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

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

đặt biểu thức đó là X

ta có :

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

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

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

\(\text{1, }\frac{27^4.9^3}{81^2}=\frac{\left(3^3\right)^4.\left(3^2\right)^3}{\left(3^4\right)^2}=\frac{3^{12}.3^6}{3^8}=3^{10}\)

\(\text{2, }\left(\frac{1}{5}\right)^{2002}.\left(-5\right)^{2000}=\frac{1}{5^{2002}}.5^{2000}=\frac{5^{2000}}{5^{2002}}=\frac{1}{5^2}=\frac{1}{5^2}\)

\(\text{3, }\frac{4^{11}.4^5}{2^{31}}=\frac{2^{22}.2^{10}}{2^{31}}=\frac{2^{32}}{2^{31}}=2\)

\(\text{4, }3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.81^2}{3^5.3^2}=\frac{3^2.3^8}{3^7}=\frac{3^{10}}{3^7}=3^3=27\)

\(\text{5, }4^2.\frac{25^2}{2^3.5^2}+32.125=\frac{2^4.5^4}{2^3.5^2}+2^5.5^3=2.5^2+2^5.5^2=5^2.\left(2+2^5.5\right)=25.\left(2+32.5\right)=25.162=4050\)