Rút gọn biểu thức B.

Điều kiện:$x\ne -5;x\ne \pm 3.$

$B=\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+9}{x^2-9}$

$\begin{array}{l}=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ =\frac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ =\frac{3x-9}{\left(x-3\right)\left(x+3\right)}=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\frac{3}{x+3}.\end{array}$

Vậy $B=\frac{3}{x+3}$  với $x\ne -5;x\ne \pm 3.$