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

a) Ta có : \(3A=3^{2007}+3^{2006}+...+3^3+3^2\)

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

\(\Rightarrow2A=3^{2007}-3\)

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

b) Ta có \(2A=3^{2007}-3\)\(\Rightarrow2A+3=3^{2007}\)

Theo bài ta có: \(2A+3=3x\)

\(\Rightarrow3^{2007}=3x\)

\(\Rightarrow3.3^{2006}=3x\)

\(\Rightarrow x=3^{2006}\)

Câu hỏi của Yuki Yudai - Toán lớp 6 - Học toán với OnlineMath

Em tham khảo nhé!

Ta có:

      A=\(3^1+3^2+....+3^{2006}\)

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

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

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

=>2A+3=\(3^x\)

<=>\(3^{2007}-3+3\)=\(3^x\)

<=>\(3^{2007}=3^x\)

=>x=2007

Vậy x=2007 thì...

dễ này mình giai cho

tính; A= 3^1+3^2 +....+3^2006 

      3A =3^2+3^3+....+3^2007

      3A-A =2A=[3^2+3^3+...+3^2007]-[3^1+3^2+...+3^2006]

      2A=3^2007-3^1

Vậy 2A+3=3^x

      3^2007-3+3=3^x

Nên 3^x = 3^2007

     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+

1/ 3A-A=3²⁰⁰⁷-3 <=> 2A=3²⁰⁰⁷-3 => A=\(\frac{3^{2007}-3}{2}\)

2/ 2A=3²⁰⁰⁷-3 => 2A+3=3²⁰⁰⁷=3^x => x=2007