Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: A = 1 + 2 + 22 + 23 + ....... + 2200
=> 2A = 2 + 22 + 23 + ....... + 2201
=> 2A - A = ( 2 + 22 + 23 + ....... + 2201 ) - ( 1 + 2 + 22 + 23 + ....... + 2200 )
=> A = 2201 - 1
=> A + 1 = 2201
A = 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 200
2A = 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + ... + 2 ^ 201
2A - A = ( 2 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4 + ... + 2 ^ 201 )
- ( 1 + 2 + 2 ^ 2 + 2 ^ 3 + ... + 2 ^ 200 )
A = 2 ^ 201 - 1
=> A + 1 = 2 ^ 201
B = 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2005
3B = 3 ^ 2 + 3 ^ 3 + 3 ^ 4 + ... + 3 ^ 2006
3B - B = ( 3 ^ 2 + 3 ^ 3 + 3 ^ 4 + ... + 3 ^ 2006 )
- ( 3 + 3 ^ 2 + 3 ^ 3 + ... + 3 ^ 2005 )
2B = 3 ^ 2006 - 3
=> 2B = 3 ^ 2006
Vậy 2B + 3 là lũy thừa của 3
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)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)
\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)
\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)
\(0,001=\frac{1}{1000}=\frac{1}{10^3}=10^{-3}\)
\(0,0001=\frac{1}{10000}=\frac{1}{10^4}=10^{-4}\)
\(0,00015=\frac{3}{20000}=\frac{3}{2}\times\frac{1}{10000}=\frac{3}{2}\times\frac{1}{10^4}=\frac{3}{2}\times10^{-4}\)
\(5^{-a}=\frac{1}{5^a}\)
\(3,5\times10^{-5}=3,5\times\frac{1}{10^5}\)
\(\left(\frac{2}{3}\right)^{-2}==\frac{1}{\left(\frac{2}{3}\right)^2}=\left(\frac{3}{2}\right)^2\)
\(10^{-3}=\frac{1}{10^3}=\frac{1}{1000}\)
2 Viết dưới dạng luỹ thừa
a) \(-729=\left(-9\right)^3.\)
b) \(-64=\left(-4\right)^3.\)
c) \(-125=\left(-5\right)^3.\)
d) \(625=25^2=\left(-25\right)^2=5^4=\left(-5\right)^4.\)
e) \(256=16^2=\left(-16\right)^2.\)
f) \(196=14^2=\left(-14\right)^2.\)
g) \(169=13^2=\left(-13\right)^2.\)
h) \(121=11^2=\left(-11\right)^2.\)
i) \(144=12^2=\left(-12\right)^2.\)
Chúc bạn học tốt
1,
4339-1737=4338.43-1736.17
=(...9)19.43-(...9)18.17
=(...9).43-(...1).17
=(...7)-(...7)=(...0) ⋮ 10 (vì chữ số tận cùng là 0)
2,
-729= -93
-64= -43
-125= -53
625= 54= -54
256= 162= -162
196= 142= -142
a) 272 : 253
= (33)2 : (52)3
= 36 : 56
\(=\left(\frac{3}{5}\right)^6\)
b) 254 : 28
= (52)4 : 28
= 58 : 28
\(=\left(\frac{5}{2}\right)^8\)
\(2T=2^2+2^3+2^4+...+2^{2009}\)
\(T=2T-T=2^{2009}-2=2\left(2^{2008}-1\right)\)
T= 2+22+23+...+22008
2T=22+23+24+...+22009
2T-T= 22009-2
T= 22009-2 = (22009-2)1
Câu 1:
2A=2+22+...+2201
A=2A-A=2201-1
⇒A+1=2201 là một lũy thừa.
Câu 2:
3B=32+33+...+32006
2B=3B-B=32006-3
⇒2B+3=32006 là một lũy thừa của 3(ĐPCM)
Câu 3 không rõ đề nhé!
\(A=1+2+2^2+...+2^{200}\)
\(\Rightarrow2A=2+2^2+...+2^{201}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{201}\right)-\left(1+2+2^2+...+2^{200}\right)\)
\(\Rightarrow A=2^{201}-1\)
\(\Rightarrow A+1=2^{201}\)