Question

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

Answer 1

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