Answer
$\frac{x+9}{(x-1)(x-3)(x+1)}$
Work Step by Step
Step 1. Factor the denominators, we have $x^2-4x+3=(x-1)(x-3)$ and $x^1-1=(x-1)(x+1)$, thus the Least Common Denominator (LCD) can be found as $(x-1)(x-3)(x+1)$
Step 2. Convert each term with the LCD, we have $\frac{3}{x^2-4x+3}-\frac{2}{x^1-1}=\frac{3(x+1)}{(x-1)(x-3)(x+1)}-\frac{2(x-3)}{(x-1)(x-3)(x+1)}=\frac{3x+3-2x+6}{(x-1)(x-3)(x+1)}=\frac{x+9}{(x-1)(x-3)(x+1)}$
