Tổng quát:

S= \(x^0+x^1+x^2+....+x^n\)

<=> x.S =x( \(x^0+x^1+x^2+....+x^n\))

<=> x.S = \(x.x^0+x.x^1+x.x^2+....+x.x^n\)

<=> x.S= \(x^1+x^2+x^3+....+x^n+x^{n+1}\)

<=> 1+x.S= 1+ \(x^1+x^2+x^3+....+x^n+x^{n+1}\) = S + \(x^{n+1}\)

=> (x-1).S = \(x^{n+1}\) -1

=> S = \(\frac{x^{n+1}-1}{x-1}\)

Từ đó: A = \(1+\frac{5^9}{1+5+5^2+...+5^8}\) = \(1+\frac{5^9\left(5-1\right)}{5^9-1}\)=\(1+\frac{5^94-4+4}{5^9-1}\)=5+\(\frac{4}{5^9-1}\)

Tương tự B= 3+ \(\frac{2}{3^9-1}\)

Vậy A > B

d. (3x-\(\frac{1}{2}\))\(^3\) = -\(\frac{1}{9}:3=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)

=> \(3x-\frac{1}{2}=\frac{-1}{3}\)

=> 3x = \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

=> x= \(\frac{1}{18}\)