Answer:
[tex]\frac{x+3}{x}[/tex]
Step-by-step explanation:
Given
[tex]\frac{x^2-9}{x^2-3x}[/tex] factorise numerator and denominator
x² - 9 = (x - 3)(x + 3) ← difference of squares
x² - 3x = x(x - 3) , thus
= [tex]\frac{(x-3)(x+3)}{x(x-3)}[/tex] ← cancel (x - 3) on numerator/ denominator, leaving
= [tex]\frac{x+3}{x}[/tex]
