Answer:
[tex]\dfrac{x+3}{x}[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2-9}{x^2-3x}\\\\=\dfrac{x^2-3^2}{x(x-3)}\\\\\text{Apply }a^2-b^2=(a-b)(a+b)\text{ formula on the numerator:}\\\\\dfrac{(x-3)(x+3)}{x(x-3)}\\\\\text{Cancel }(x-3)\text{ from numerator and denominator:}[/tex]
[tex]\dfrac{x+3}{x}\\\\\therefore\ \dfrac{x^2-9}{x^2-3x}=\dfrac{x+3}{x}[/tex]
