Respuesta :

caylus

Answer:[tex]15*(x-3)^2*(x-5)[/tex]

Step-by-step explanation:

[tex]\dfrac{x^2-y^2}{x^2-8x+15}\ and\ \dfrac{y^2}{15(x^2-6x+9)} \\\\\\x^2-8x+15=x^2-3x-5x+15=x(x-3)-5(x-3)=(x-3)(x-5)\\\\15(x^2-6x+9)=15*(x-3)^2\\\\LCM (x^2-8x+15,15(x^2-6x+9)\\\\=LCM((x-3)(x-5),15(x-3)^2)\\\\\boxed{=15*(x-3)^2*(x-5)}\\[/tex]