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

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Hok tốt :>>

-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=3+3²+3³+...+3²⁰¹⁹

3A=3²+3³+...+3²⁰²⁰

3A-A=(3²+3³+...+3²⁰²⁰)-(3+3²+3³+...+3²⁰¹⁹)

2A=3²⁰²⁰-3

2A+3=3²⁰²⁰

⇒n=2020

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 3^x=3²⁰⁰⁷

^=>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

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