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=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

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+

\(A=3+3^2+3^3+.......+3^{2006}\)

\(\Leftrightarrow3A=3^2+3^3+......+3^{2007}\)

\(\Leftrightarrow3A-A=3^{2007}-3\)

\(\Leftrightarrow2A=3^{2007}-3\)

\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)

\(\Leftrightarrow2A+3=2^{2007}\)

\(\Leftrightarrow2^{2007}=2^x\)

\(\Leftrightarrow x=2007\)

\(3A=3^2+3^3+....+3^{2007}\)

\(3A-A=\left(3^2+3^3+...+3^{2007}\right)-\left(3+3^2+...+3^{2006}\right)\)

\(2A=3^{2007}-3\)

\(A=\frac{3^{2007}-3}{2}\)

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=3^x

thay 2A=3²⁰⁰⁷-3 vào ta được

<=>3²⁰⁰⁷-3+3=3^x

<=>3²⁰⁰⁷=3^x

<=>x=2007

\(3A=3^2+3^3+3^4+...+3^{2007}\)

\(3A-A=2A=3^{2007}-3\)

\(A=\frac{3^{2007}-3}{2}\)

\(3A=3^2+3^3+3^4+...+3^{2007}\)

\(3A-A=2A=3^{2007}-3\)

\(A=\frac{3^{2007}-3}{2}\)

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) \(2A+3=-3+3-3^{2017}=3^{2017}=3^x\)

\(\Rightarrow x=2017\)

mình cũng ko biết câu này

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

\(A=3+3^2+3^3+.........+3^{2006}\)

\(\Leftrightarrow3A=3^2+3^3+.........+3^{2007}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{2007}\right)-\left(3+3^2+.....+3^{2006}\right)\)

\(\Leftrightarrow2A=3^{2007}-3\)

\(\Leftrightarrow A=\frac{3^{2007}-3}{2}\)

\(\Leftrightarrow2A+3=3^{2007}\)

\(\Leftrightarrow3^x=3^{2007}\)

\(\Leftrightarrow x=2007\left(tm\right)\)

a)\(A=3+3^2+3^3+...+3^{2006}\)

=>\(3A=3\left(3+3^2+3^3+...+3^{2006}\right)=3^2+3^3+3^4+...+3^{2007}\)

=>\(3A-A=\left(3^2+3^3+3^4+...+3^{2007}\right)-\left(3+3^2+3^3+...+3^{2006}\right)\)

=>\(2A=3^{2007}-3\Rightarrow A=\frac{3^{2007}-3}{2}\)

b)\(2A+3=3^x\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\Rightarrow3^{2007}=3^x\Rightarrow x=2007\)

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 = 3^x

=> 3^x = 3²⁰⁰⁷

=> x = 2007