`a)đk:a>0,a ne 9`

`A=((sqrta+3+sqrta-3)/(a-9)).((sqrta-3)/sqrta)`

`=((2sqrtx)/(a-9)).((sqrta-3)/sqrta)`

`=2/(sqrta+3)`

`b)A>1/2`

`<=>2/(sqrta+3)>1/2`

`<=>sqrta+3<4`

`<=>sqrta<1`

`<=>a<1`

KẾt hợp đkxđ:`0<x<1`

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne9\end{matrix}\right.\)

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-3}+\dfrac{1}{\sqrt{a}+3}\right)\left(1-\dfrac{3}{\sqrt{a}}\right)\)

\(=\dfrac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\dfrac{\sqrt{a}-3}{\sqrt{a}}\)

\(=\dfrac{2\sqrt{a}}{\sqrt{a}+3}\cdot\dfrac{1}{\sqrt{a}}\)

\(=\dfrac{2}{\sqrt{a}+3}\)

b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{2}{\sqrt{a}+3}-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{4-\left(\sqrt{a}+3\right)}{2\left(\sqrt{a}+3\right)}>0\)

mà \(2\left(\sqrt{a}+3\right)>0\forall a\)

nên \(1-\sqrt{a}>0\)

\(\Leftrightarrow\sqrt{a}< 1\)

hay a<1

Kết hợp ĐKXĐ, ta được: 0<a<1

Theo công thức biến đổi có số ta có : \(\log_{a^n}x=\frac{\log_ax}{\log_aa^n}=\frac{1}{n}\log_ax\)

Từ đó ta có :

      \(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}\)

          \(=\frac{1}{\log_ax}+\frac{2}{\log_ax}+\frac{4}{\log_ax}+...+\frac{n}{\log_ax}\)

          \(=\frac{1+2+3+...+n}{\log_ax}=\frac{n\left(n+1\right)}{\log_ax}\)

Vậy \(A=\frac{1}{\log_ax}+\frac{1}{\log_{a^2}x}+\frac{1}{\log_{a^3}x}+...+\frac{1}{\log_{a^n}x}=\frac{n\left(n+1\right)}{\log_ax}\)

b,     B        =                       \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\)  + \(\dfrac{1}{2^3}\) -   \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)

2 \(\times\)  B       =                 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) -  \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)

2 \(\times\) B + B  =                1  -  \(\dfrac{1}{2^{100}}\)

3B             =              ( 1 - \(\dfrac{1}{2^{100}}\)) 

             B =               ( 1 - \(\dfrac{1}{2^{100}}\)) : 3

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

A\(\times\)  3             =   3 +  1 + \(\dfrac{1}{3}\) +  \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+  \(\dfrac{1}{3^{n-1}}\) 

A \(\times\) 3 - A        = 3 - \(\dfrac{1}{3^n}\)

       2A           = 3  - \(\dfrac{1}{3^n}\)

         A           = ( 3 - \(\dfrac{1}{3^n}\)) : 2

C = \(\dfrac{3}{2^2}\) \(\times\) \(\dfrac{8}{3^2}\) \(\times\) \(\dfrac{15}{4^2}\) \(\times\) ...........\(\times\) \(\dfrac{899}{30^2}\)

C = \(\dfrac{1\times3}{2^2}\) \(\times\) \(\dfrac{2\times4}{3^2}\) \(\times\) \(\dfrac{3\times5}{4^2}\) \(\times\)........\(\times\) \(\dfrac{29\times31}{30^2}\)

C = \(\dfrac{1\times2\times\left(3\times4\times5\times....\times29\right)^2\times30\times31}{2^2\times\left(3\times4\times5\times.......\times29\right)^2\times30^2}\)

C =  \(\dfrac{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}{2\times\left(3\times4\times5\times.....\times29\right)^2\times30}\) \(\times\) \(\dfrac{1\times31}{2\times30}\)

C = 1 \(\times\) \(\dfrac{31}{60}\)

C = \(\dfrac{31}{60}\)

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

a) \(A=1+3+3^2+...+3^{100}\)

\(3A=3+3^2+3^3+...+3^{101}\)

\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2A=3^{101}-1\)

\(A=\frac{3^{101}-1}{2}\)

b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)

\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)

\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)

\(3B=2^{101}+1\)

\(B=\frac{2^{101}+1}{3}\)

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

\(A=2A-A=2-\frac{1}{2^{2011}}=\frac{2^{2012}-1}{2^{2011}}\)

Nhầm

\(A=2A-A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
 

=5x^2+5x-2x-2-(5x^2+x-15x-3)-17x-51

=5x^2-14x-53-5x^2+14x+3

=-50