CreolaTeti359
19

# Need a step by step on how to solve this please

$\frac{\frac{1}{(x+3)^2}-\frac{1}{x^2}}{3}=\frac{1}{3(x+3)^2}-\frac{1}{3x^2}=\frac{x^2}{3x^2(x+3)^2}-\frac{(x+3)^2}{3x^2(x+3)^2}=\frac{x^2-x^2-6x-9}{3x^3(x+3)^2}\\\\=\frac{-6x-9}{3x^2(x+3)^2}=\frac{3(-2x-3)}{3x^2(x+3)^2}=-\frac{2x+3}{x^2(x+3)^2}$