\(\frac{x-1}{x-3}\)
\(\frac{-2}{x+3}\)
\(\frac{x-1}{x+3}\)
\(\frac{4x}{x^2-9}\)
Correct answer is B
\(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}\\
\frac{1}{x-3}-\frac{3(x-1)}{(x-3)(x+3)}\\
\frac{x+3-3x+3}{(x-3)(x+3)};\frac{-2x+6}{(x-3)(x+3)}\\
\frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3}\)