Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có
\(3A=3^2+3^3+....+3^{201}\)
\(\Rightarrow3A-A=2A=\left(3^2+3^3+....+3^{201}\right)-\left(3+3^2+....+3^{200}\right)\)
\(\Rightarrow2A=3^{201}-3\)
\(\Rightarrow2A-3=3^{201}\)
Mà 2A - 3= 3x
=>x=3200
3a=32+33+...+3201
3a-a=2a= (3^2+3^3+...+3^201)-(3+3^2+...+3^201)
2a =3^201 -3
2a-3 =3^201
=>x= 3^200
\(A=3+3^2+3^3+.......+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+......+3^{2007}\)
\(\Leftrightarrow3A-A=3^{2007}-3\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
\(\Leftrightarrow2A+3=2^{2007}\)
\(\Leftrightarrow2^{2007}=2^x\)
\(\Leftrightarrow x=2007\)
\(3A=3^2+3^3+....+3^{2007}\)
\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3+3^2+...+3^{2006}\right)\)
\(2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
b)\(2A+3=3^x\)
\(2A=3^x-3\)
Mà:\(2A=3^{2007}-3\)
\(\Rightarrow x=2007\)
A = 31+32+33+.....+32006
3A = 32+33+34+....+32007
2A = 3A - A = 32007-3
=> A = \(\frac{3^{2007}-3}{2}\)
Vì 2A = 32007-3
=> 2A + 3 = 32007
Mà 2A + 3 = 3x
=> 3x = 32007
=> x = 2007
1/ 3A-A=32007-3 <=> 2A=32007-3 => A=\(\frac{3^{2007}-3}{2}\)
2/ 2A=32007-3 => 2A+3=32007=3x => x=2007
a ) A = 3 + 32 + 33 + .... + 32012
Nhan của 2 vế của A với 3 ta được :
3A = 3(3 + 32 + 33 + .... + 32012)
= 32 + 33 + 34 + .... + 32013
Trừ cả hai vế của 3A cho A ta được :
3A - A = (32 + 33 + 34 + .... + 32013) - (3 + 32 + 33 + .... + 32012)
2A = 32013 - 3
=> A = (32013 - 3) : 2
b ) Theo a ) ta có :
2A = 32013 - 3 => 2A + 3 = 32013
Mà theo đề bài : 2A + 3 = 3x
=> 32013 = 3x => x = 2013
Vậy x = 2013
Ta có : \(A=3+3^2+3^3+......+3^{2006}\)
=> \(3A=3^2+3^3+......+3^{2007}\)
=> \(3A-A=3^{2007}-3\)
=> \(2A=3^{2007}-3\)
=> \(A=\frac{3^{2007}-3}{2}\)
b) Ta có : \(2A=3^{2007}-3\) (theo ý a)
=> \(2A+3=3^{2007}\)
=> x = 2007
\(A=3+3^2+3^3+.........+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+.........+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{2007}\right)-\left(3+3^2+.....+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
\(\Leftrightarrow2A+3=3^{2007}\)
\(\Leftrightarrow3^x=3^{2007}\)
\(\Leftrightarrow x=2007\left(tm\right)\)
3A=\(3^2+3^3+3^4+...+3^{2007}\)
3A-A=2A=\(3^{2007}-3\)
A=\(\frac{3^{2007}-3}{2}\)
b.
2A+3=3^x
3^2007-3+3=3^x
3^2007=3^x
vay x=2007
ta có : 3A=32+33+...+32007
3A-A=32+33+34+....+32007-3-32-33-...-32006
2A=32007-3
A=\(\frac{3^{2007}-3}{2}\)
b,
2A+3=3x
<=>32007-3+3=3x
<=> 32007=32007
<=> x = 2007
vậy x =2007
Ta có: 5^200=5^100.5^100=(5.5)^100=25^100
:3^300=3^100.3^100.3^100=(3.3.3)^100=27^100
Vì 27>25 => 25^100<27^100 Hay 5^200<3^300
\(A=3+3^2+3^3+...3^{2006}\)
\(3A=3^2+3^3+...+3^{2007}\)
\(3A-A=\left(3^2-3^2\right)+....+\left(3^{2006}-3^{2006}\right)+3^{2007}-3\)
\(2A=3^{2007}-3\Rightarrow2A+3=3^{2007}-3+3=3^{2007}=3^x\)
Vậy x = 2007
A=3+3^2+....+3^2006
=>3A=3^2+3^3+....+3^2007
=>3A-A=(3^2+3^3+....+3^2007)-(3+3^2+....+3^2006)
=>2A=3^2007-3
khi đó 2A+3=3^2007-3+3=3^2007=3^x
=>x=2007
1.
a) A = 3^1 + 3^2 +........+3^2006
3A = 3^2 + ............+3^2006 + 3^2007
3A - A = (3^2 +........+3^2006 +3^2007)-(3^1 + 3^2+.....+3^2006)
2A = 3^2007 - 3^1
Sửa đề: \(A=3+3^2+3^3+\cdots+3^{299}\)
=>\(3A=3^2+3^3+\cdots+3^{300}\)
=>3A-A=\(3^2+3^3+\cdots+3^{300}-3-3^2-\cdots-3^{299}\)
=>2A=\(3^{300}-3\)
=>2A+3=\(3^{300}\)
=>\(3^{300}-x=3^{300}\)
=>x=0
🔞🔞🔞
3A=3^2+3^3+...+3^2007
=>3a-A=(3^2+3^3+...+3^2007)-(3^1+3^2+...+3^2006)
=>2A=3^2007-3^1=3^2007-3
=>2A+3=3^2007-3+3=3^2007=3^x
=>x=2007