Cho A=3^1+3^2+3^3+...+3^2006
Thu gọn A
Tìm x để 2A+3=3^x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: \(x\notin\left\{-3;3\right\}\)
b: \(P=\dfrac{x^2-2x-3-x^2-5x-6+4x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-3x-3}{\left(x-3\right)\left(x+3\right)}\)
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
-Ta có:1+2+3+.........+2006=(2006+1).2006:2=2013021
A=31+
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, A=31+32+33+...+32006
3A=32+33+...+32006+32007
3A-A=(32+33+...+32006+32007)-(31+32+33+...+32006)
2A=32007-3
A=(32007-3)/2
b, 2A=32007-3
2A+3=32007
Hay 3x=32007
=>x=2007
\(3A=3^2+3^3+3^4+...+3^{2007}\)
\(\Rightarrow3A-A=2A=3^{2007}-3^1=3.\left(3^{2006}-1\right)\)
Do đó \(A=\frac{3.\left(3^{2006}-1\right)}{2}\)
Ta có : \(2A+3=3^{2007}-3+3=3^{2007}=3^x\)
\(\Rightarrow x=2007\)
\(A=3+3^2+3^3+...+3^{2006}\)
\(\Leftrightarrow3A=3\left(3+3^2+3^3+....+3^{2006}\right)\)
\(\Leftrightarrow3A=3^2+3^3+3^4+....+3^{2007}\)
\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)
\(\Leftrightarrow2A=3^{2007}-3\)
\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)
Ta có \(2A=3^{2007}-3\)
=> 2A+3=\(3^{2007}-3+3=3^{2007}\)
=> x=2007
A=3^1+3^2+3^3+....+3^2006
3A=3^2+3^3+...+3^2007
=>2A=3^2007-3
=>2A+3=3^x
3^2007-3+3=3^x
3^2007=3^x
=>x=2007
Vậy x=2007