try to factor them and cancel factors
x-3 is factored
[tex]x^2-25=x^2-5^2=(x+5)(x-5)[/tex]
[tex]x^2-9=x^2-3^2=(x+3)(x-3)[/tex]
x-5 is factored
also remember that
[tex]\frac{\frac{a}{b}}{\frac{c}{d}}=(\frac{a}{b})(\frac{d}{c})=[/tex] [tex]\frac{ad}{bc}[/tex]
therefor
[tex]\frac{\frac{x-3}{x^2-25}}{\frac{x^2-9}{x-5}}=[/tex]
[tex]\frac{\frac{x-3}{(x+5)(x-5)}}{\frac{(x+3)(x-3)}{x-5}}=[/tex]
[tex](\frac{x-3}{(x+5)(x-5)})(\frac{x-5}{(x+3)(x-3)}=[/tex]
[tex]\frac{(x-3)(x-5)}{(x+5)(x-5)(x+3)(x-3)}=[/tex]
[tex](\frac{1}{(x+5)(x+3)})(\frac{(x-3)(x-5)}{(x-3)(x-5)}=[/tex]
[tex](\frac{1}{x^2+8x+15})(1)=[/tex]
[tex]\frac{1}{x^2+8x+15}[/tex]
