Cho A = 1 + 3 + 3 mũ 2 + ...... + 3 mũ 2007 hãy viết 2A dưới dạng 1 luỹ thừa
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ND
1
7 tháng 8 2016
\(A=1+3+3^2+...+3^{2007}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2008}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2008}\right)-\left(1+3+3^2+...+3^{2007}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{2008}-1-3-3^2-...-3^{2007}\)
\(\Rightarrow2A=3^{2008}-1\)
\(\Rightarrow2A+1=3^{2008}\)
ND
1
7 tháng 8 2016
\(A=1+3+3^2+...+3^{2007}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2008}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2008}\right)-\left(1+3+3^2+...+3^{2007}\right)\)
\(\Rightarrow2A=3+3^2+3^3+...+3^{2008}-1-3-3^2-...-3^{2007}\)
\(\Rightarrow2A=3^{2008}-1\)
\(\Rightarrow2A+1=3^{2008}\)
Nhớ k cho mk nha!!!
30 tháng 9 2018
a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)
b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)
c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)
T
30 tháng 9 2018
\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)
\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)
1 tháng 9 2021
a: \(4^5\cdot8^7=2^{10}\cdot2^{21}=2^{31}\)
b: \(125^5\cdot25^3=5^{15}\cdot5^6=5^{21}\)
PA
9 tháng 7 2016
\(0,001=\frac{1}{1000}=\frac{1}{10^3}=10^{-3}\)
\(0,0001=\frac{1}{10000}=\frac{1}{10^4}=10^{-4}\)
\(0,00015=\frac{3}{20000}=\frac{3}{2}\times\frac{1}{10000}=\frac{3}{2}\times\frac{1}{10^4}=\frac{3}{2}\times10^{-4}\)
\(5^{-a}=\frac{1}{5^a}\)
\(3,5\times10^{-5}=3,5\times\frac{1}{10^5}\)
\(\left(\frac{2}{3}\right)^{-2}==\frac{1}{\left(\frac{2}{3}\right)^2}=\left(\frac{3}{2}\right)^2\)
\(10^{-3}=\frac{1}{10^3}=\frac{1}{1000}\)
VL
3
10 tháng 10 2017
3A=\(3+3^2+3^3+...+3^{11}\)
3A-A=(\(3+3^2+3^3+...+3^{11}\))-(\(1+3+3^2+...+3^{10}\))
2A=\(3^{11}-1\)
2A+1=\(3^{11}\)
3A=3+32+33+....+32008
2A=(3+32+....+32008)-(1+3+...+32007)=32008-1