Решите неравенство x^2-4x+3/x-2 - x-3/x^2-3x+2<=0

$\frac{x^2-4x+3}{x-2} - \frac{x-3}{x^2-3x+2} \leq 0\\\\ \frac{(x-1)(x-3)}{x-2} - \frac{x-3}{(x-1)(x-2)} \leq 0\\\\ \frac{(x-1)^2(x-3)-(x-3)}{(x-1)(x-2)} \leq 0\\\\ \frac{(x-3)((x-1)^2-1)}{(x-1)(x-2)} \leq 0\\\\ \frac{(x-3)(x-1-1)(x-1+1)}{(x-1)(x-2)} \leq 0\\\\ \frac{(x-3)(x-2)x}{(x-1)(x-2)} \leq 0$
- + - - +
_______________(0)___________(1)__________(2)_____________[3]______________

x∈(-∞;0)U(1;2)U(2;3]