Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(A=3+3^2+3^3+...+3^{2006}\)
\(\Leftrightarrow3A=3\left(3+3^2+3^3+....+3^{2006}\right)\)
\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
Ta có \(2A=3^{2007}-3\)
=> 2A+3=\(3^{2007}-3+3=3^{2007}\)
=> x=2007
a,Ta có:3A=32+33+................+32011
\(\Rightarrow3A-A=\left(3^2+3^3+.....+3^{2011}\right)-\left(3+3^2+.....+3^{2010}\right)\)
\(\Rightarrow2A=3^{2011}-3\)
\(\Rightarrow A=\frac{3^{2011}-3}{2}\)
b,Ta có:\(2A=3^{2011}-3\Rightarrow2A+3=3^{2011}\Rightarrow x=2011\)
\(A=3+3^2+3^3+.......+3^{2006}\)
\(\Leftrightarrow3A=3^2+3^3+......+3^{2007}\)
\(\Leftrightarrow3A-A=3^{2007}-3\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
\(\Leftrightarrow2A+3=2^{2007}\)
\(\Leftrightarrow2^{2007}=2^x\)
\(\Leftrightarrow x=2007\)
\(3A=3^2+3^3+....+3^{2007}\)
\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3+3^2+...+3^{2006}\right)\)
\(2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
b)\(2A+3=3^x\)
\(2A=3^x-3\)
Mà:\(2A=3^{2007}-3\)
\(\Rightarrow x=2007\)
a)3A=3(31 + 32 + 33 + ... + 32006)
3A=32+33+...+32007
3A-A=(32+33+...+32007)-(31 + 32 + 33 + ... + 32006)
2A=32007-3
A=\(\frac{3^{2007}-3}{2}\)
b)2A+3=3x
thay 2A=32007-3 vào ta được
<=>32007-3+3=3x
<=>32007=3x
<=>x=2007
\(3A=3^2+3^3+3^4+...+3^{2007}\)
\(3A-A=2A=3^{2007}-3\)
\(A=\frac{3^{2007}-3}{2}\)
\(3A=3^2+3^3+3^4+...+3^{2007}\)
\(\Rightarrow3A-A=2A=3^{2007}-3^1=3.\left(3^{2006}-1\right)\)
Do đó \(A=\frac{3.\left(3^{2006}-1\right)}{2}\)
Ta có : \(2A+3=3^{2007}-3+3=3^{2007}=3^x\)
\(\Rightarrow x=2007\)
3A - A = (32 + 33 + 34 + ... + 32007) - (3 + 32 + 33 + ... + 32006)
2A = 32007 - 3\(\Rightarrow\hept{\begin{cases}A=\frac{3^{2007}-3}{2}\\2A+3=3^{2007}\Rightarrow x=2007\end{cases}}\)
\(A=3+3^2+3^3+...+3^{2016}\)
\(\Rightarrow3A=3\left(3+3^2+3^3+...+3^{2016}\right)\)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2017}\)
\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+3^{2016}\right)\)
\(\Rightarrow2A=-3+3^{2017}\)
\(\Rightarrow A=\frac{3+3^{2017}}{2}\)
b) \(2A+3=-3+3-3^{2017}=3^{2017}=3^x\)
\(\Rightarrow x=2017\)
Ở dưới câu của bn
có câu hỏi giống vậy đó
Hok tốt :>>
-Ta có:1+2+3+.........+2006=(2006+1).2006:2=2013021
A=31+