Question

Answer 1

11: \(=\dfrac{x^2+6x+9-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{9x}=\dfrac{4}{x-3}\)

Answer 2

\(\left(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}\right).\dfrac{x+3}{9x}\\ =\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{9x}\\ =\dfrac{\left(x+3-x+3\right)\left(x+3+x-3\right)}{9x\left(x-3\right)}\\ =\dfrac{12x}{9x\left(x-3\right)}\\ =\dfrac{4}{3\left(x-3\right)}\)