a) \(A = \frac{3 x + 15}{x^{2} - 9} + \frac{1}{x + 3} - \frac{2}{x - 3}\) (với \(x \neq 3\), \(x \neq - 3\))

\(A = \frac{3 x + 15}{\left(\right. x + 3 \left.\right) \left(\right. x - 3 \left.\right)} + \frac{1}{x + 3} - \frac{2}{x - 3}\)

\(A = \frac{3 x + 15 + x - 3 - 2 x - 6}{\left(\right. x + 3 \left.\right) \left(\right. x - 3 \left.\right)}\)

\(A = \frac{2 x + 6}{\left(\right. x + 3 \left.\right) \left(\right. x - 3 \left.\right)}\)

\(A = \frac{2}{x - 3}\).

b) Để \(A = \frac{2}{3}\) thì \(\frac{2}{x - 3} = \frac{2}{3}\)

\(x - 3 = 3\)

\(x = 6\) (thỏa mãn điều kiện).

Vậy \(x = 6\) thì \(A = \frac{2}{3}\).