Ta có: A = 2 + 2² + 2³ + ... + 2²⁰⁰⁸

=> 2A = 2² + 2³ + ... + 2²⁰⁰⁹

=> 2A - A = 2²⁰⁰⁹ - 2

=> A = 2²⁰⁰⁹ - 2

Vậy A = 2²⁰⁰⁹ - 2

=> 2A = 2.(2²⁰⁰⁹ - 2)

=> 2A = 2²⁰¹⁰ - 4

=> 2A + 4 = 2²⁰¹⁰ - 4 + 4

=> 2A + 4 = 2²⁰¹⁰

=> 3x = 2010 

Sai đề

\(A=2+2^2+2^3+...+2^{2008}\)

\(2A=2^2+2^3+2^4+...+2^{2009}\)

\(2A-A=\left(2^2+2^3+2^4+...+2^{2009}\right)-\left(2+2^2+2^3+...+2^{2008}\right)\)

\(A=2^{2009}-2\)

\(\Rightarrow2.\left(2^{2009}-2\right)+4=2^{3x}\)

\(\Rightarrow2^{2010}-4+4=2^{3x}\)

\(\Rightarrow2^{2010}=2^{3x}\)

=> 3x = 2010

=> x =670

\(A=2+2^2+2^3+...+2^{2008}\)

\(\Rightarrow2A=2^2+2^3+2^4+...+2^{2009}\)

\(\Rightarrow2A-A=2^{2009}-2\)

\(\Rightarrow A=2^{2009}-2\)

\(\Rightarrow2\left(2^{2009}-2\right)+4=2^{2010}-4+4=2^{2010}=2^{3x}\)

\(\Rightarrow2010=3x\Rightarrow x=670\)

vậy

Ta có: \(A=3+3^2+3^3+...+3^{2008}\)

\(3A=3^2+3^3+3^4+...+3^{2009}\)

\(3A-A=3^{2009}-3\)

Hay \(2A=3^{2009}-3\)

\(\Rightarrow2A+3=3^x\)

\(\Rightarrow\left(3^{2009}-3\right)+3=3^x\)

\(\Rightarrow3^{2009}=3^x\)

\(\Rightarrow x=2009\)

Hok tốt nha^^

Có A=3+3²+...+3²⁰⁰⁸

=>3A=3²+3³+...+3²⁰⁰⁹

=>3A-A=2A=3²⁰⁰⁹-3

Thay 2A vào 2A+3=3^x

Ta được: 3²⁰⁰⁹-3+3=3^x

=>3²⁰⁰⁹=3^x

=>x=2009

Vậy..

Bài 1:

a)\(\begin{cases}\left(x-3\right)^2+\left(y+2\right)^2=0\\\begin{cases}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\end{cases}\)

\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+2\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}x=3\\y=-2\end{cases}\)

b) tương tự 

b) (x-12+y)²⁰⁰+(x-4-y)²⁰⁰= 0

\(\begin{cases}\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\\\begin{cases}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{cases}\end{cases}\)

\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)\(\Rightarrow\begin{cases}x+y=12\left(1\right)\\x-y=4\left(2\right)\end{cases}\)

Trừ theo vế của (1) và (2) ta được:

\(2y=8\Rightarrow y=4\)\(\Rightarrow\begin{cases}x+4=12\\x-4=4\end{cases}\)\(\Rightarrow x=8\)

Vậy x=8; y=4

Các bạn ơi chỉ cần trả lời câu 1 thôi cũng được mình làm được câu 2 rồi

Ta có :\(A=3+3^2+3^3+...+3^{2008}\)(1)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)(2)

Lấy (2) trừ đi 1 ta có :

\(\Rightarrow2A=3^{2009}-3\)

Ta lại có :

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

\(\Rightarrow3^{2009}=3^x\)

\(\Rightarrow x=2009\)

A = 3 + 3² + 3³ + ... + 3²⁰⁰⁸
3A = 3² + 3³ + 3⁴ + ... + 3²⁰⁰⁹
3A - A = ( 3² + 3³ + 3⁴ + ... + 3²⁰⁰⁹) - ( 3 + 3² + 3³ + ... + 3²⁰⁰⁸)
2A = 3²⁰⁰⁹ - 3
A = \(\frac{3^{2009}-3}{2}\)
\(2A+3=3^x\)
\(\Rightarrow\)\(\frac{3^{2009}-3}{2}\times2+3=3^x\)
\(\Rightarrow3^{2009}-3+3=3^x\)
\(\Rightarrow3^{2009}=3^x\)
\(\Rightarrow x=2009\)

Ta có:3A=3²+3³+.................+3²⁰⁰⁹

\(\Rightarrow\)3A-A=(3²+3³+...............+3²⁰⁰⁹)-(3+3²+3³+................+3²⁰⁰⁸)

\(\Rightarrow2A=3^{2009}-3\)

\(\Rightarrow2A+3=3^{2009}\Rightarrowđpcm\)

A= 3+3^2+3^3+...3^2008

=> 3A = 3^2+3^3+...3^2008+3^2009

=> 3A-A=2A= (  3^2+3^3+...3^2008+3^2009 ) - ( 3+3^2+3^3+...3^2008 )

=> 2A= 3^2009-3

Mà : 2A+3=3ⁿ

=> 3ⁿ = 3^2009-3

còn đến đây thì bạn tìm n như tìm x thôi :)

\(A=3+3^2+...+3^{2008}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2009}\)

\(\Rightarrow3A-A=3^{2009}-3\)

\(\Rightarrow2A+3=3^{2009}\)

Vậy n = 2009

\(A=3+3^2+3^3+...+3^{2008}\)

\(\Leftrightarrow3A=3^2+3^3+...+3^{2009}\)

\(\Leftrightarrow3A-A=3^{2009}-3\Leftrightarrow2A+3=3^{2009}\)

Vậy n=2009

\(A=3+3^2+3^3+...+3^{2008}\)

\(\Rightarrow3A=3\cdot\left(3+3^2+3^3+...+3^{2008}\right)\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)

\(\Rightarrow3A-A=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)

\(\Rightarrow2A=3^{2009}-3\)

Ta có: \(2A+3=3^x\)

\(\Rightarrow3^{2009}-3+3=3^x\)

\(\Rightarrow3^{2009}=3^x\)

\(\Rightarrow x=2009\)

Trả lời :

Nhân hai vế với 3 , ta được :

  \(3A=3^2+3^3+3^4+...+3^{2009}\)    ( 2 )

-   \(A=3+3^2+3^3+...+3^{2008}\)      ( 1 )

__________________________________________

\(2A=3^{2009}-3\)

Từ ( 1 ) và ( 2 ), ta có :

\(2A=3^{2009}-3\Leftrightarrow2A+3=3^{2009}\Rightarrow3^x=3^{2009}\Rightarrow x=2009\)

     - Study well -

\(A=3+3^2+...+3^{2008}\)

\(3A=3.\left(3+3^2+...+3^{2008}\right)\)

\(3A-A=\left(3^2+3^3+...+3^{2009}\right)-\left(3+3^2+...+3^{2008}\right)\)

\(2A=3^{2009}-3\)

\(2A+3=3^{2009}-3+3\)

\(2A+3=3^{2009}\)

Vì \(2A+3=3^x\)hay \(3^{2009}=3^x\)

 \(\Rightarrow x=2009\)

Thank you to kick me ooooooooooooooooooo