[tex]\frac{ z-1 }{z^2+3z+2} - \frac{z- 3 }{2z^2+3z-2 }=\frac{( z-1 )(z^2+3z-2)-(z-3)(z^2+3z+2)}{(z^2+3z+2)(z^2+3z-2)} = \\\\=\frac{ z^3+3z^2-2z -z^2-3z+2 -( z^3+3z^2+2z-3z^2-9z-6)}{(z^2+3z+2)(z^2+3z-2)} =\\\\=\frac{ z^3+3z^2-2z -z^2-3z+2 - z^3-3z^2-2z+3z^2+9z+6 }{(z^2+3z+2)(z^2+3z-2)} =\\\\= \frac{2z^2 +2z +8 }{(z^2+3z+2)(z^2+3z-2)} = \frac{2(z^2 + z +4) }{(z^2+3z+2)(z^2+3z-2)}[/tex]
[tex]z^2+3z+2 \neq 0 \ \ \wedge \ \ 2z^2+3z-2\neq 0 \\\\(z+1)(z+2)\neq 0 \ \ \wedge \ \ (z+2)(2z-1)\neq 0\\\\ z+1 \neq 0 \ \ \wedge \ \ z+2 \neq 0 \ \ \wedge \ \ z+2 \neq 0\ \ \wedge 2z-1 \neq 0\\\\z \neq -1 \ \ \wedge \ \ z \neq -2\ \ \wedge \ \ z \neq -2\ \ \wedge z \neq \frac{1}{2} \\\\D=R\setminus \left \{ -1,-2,\frac{1}{2} \right \}[/tex]
