$\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}$
$=\frac{1}{x^2-2x-x+2}+\frac{1}{x^2-2x-3x+6}-\frac{2}{x^2-3x-x+3}$
$=\frac{1}{x(x-2)-(x-2)}+\frac{1}{x(x-2)-3(x-2)}-\frac{2}{x(x-3)-(x-3)}$
$=\frac{1}{(x-1)(x-2)}+\frac{1}{(x-3)(x-2)}-\frac{2}{(x-1)(x-3)}$
$=\frac{1(x-3)+1(x-1)-2(x-2)}{(x-1)(x-2)(x-3)}$
$=\frac{x-3+x-1-2x+4}{(x-1)(x-2)(x-3)}$
$=\frac{0}{(x-1)(x-2)(x-3)}=0$
