[tex]x+2\neq 0\ \ lub\ \ x-2\neq 0 \\\\ x \neq -2\ \ lub\ \ x \neq 2\\\\D=R\ \setminus \left\{-2,2 \right\}[/tex]
[tex]\frac{x^2-9}{x^2-4}+\frac{x-3}{x+2}-\frac{1-x}{x-2}=\frac{x^2-9}{(x+2)(x-2) }+\frac{x-3}{x+2}-\frac{1-x}{x-2}= \frac{x^2-9+(x-3) (x-2)-(1-x)(x+2)}{(x+2)(x-2) } =\\\\\\=\frac{x^2-9+x^2-2x-3x+6-(x+2-x^2-2x)}{(x+2)(x-2) } =\frac{2x^2-3 -5 x - x-2+x^2+2x }{(x+2)(x-2) }=\\\\= \frac{3x^2-4x-5}{(x+2)(x-2) }[/tex]
