x khac +-3
A=\(\hept{\begin{cases}\\\end{cases}\frac{xI\left(x-3\right)I}{5x^2-45}=\frac{xI\left(x-3\right)I}{5\left(x^2-3^2\right)}}\)
\(\frac{xIx-3I\overline{ }}{5\left(x-3\right)\left(x+3\right)^{ }_{ }}\)
x>3 A=\(\frac{x}{5\left(x+3\right)}\)
x<3 A=-\(\frac{x}{5\left(x+3\right)}\)