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+

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

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

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

\(\Rightarrow3A-A=2A=3^{2007}-3^1=3.\left(3^{2006}-1\right)\)

Do đó \(A=\frac{3.\left(3^{2006}-1\right)}{2}\)

Ta có : \(2A+3=3^{2007}-3+3=3^{2007}=3^x\)

\(\Rightarrow x=2007\)

x=

2007

đúng ko

vậy hả

bn

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

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

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

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

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

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

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

\(\Leftrightarrow x=2007\)

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

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

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

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

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

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

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

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

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

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

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