## anonymous 5 years ago (3x-9)/[12(x-(9/x))]

$\frac{3x-9}{12(x-9/x)}=\frac{3x^2-9x}{12(x^2-9)}=\frac{3x(x-3)}{12(x+3)(x-3)}=\frac{x}{4(x+3)}$ First multiply the top and bottom by x, then factor, then reduce. if you haven't seen the difference of squares factoring before, which I used in the second step, it's very useful. $a^2-b^2=(a+b)(a-b)$