Question

(x+$frac{9}{x-6})$÷$frac{3x^{2}-18x+27}{x^{2}-96}$

Answer 1

$(x+frac{9}{x-6}):frac{3x^2-18x+27}{x^2-36}=frac{x(x-6)+9}{x-6}:frac{3cdot (x^2-6x+9)}{(x-6)(x+6)}=\\=frac{x^2-6x+9}{x-6}cdot frac{(x-6)(x+6)}{3(x^2-6x+9)}=frac{x+6}{3}$