ĐK: $x\ne \pm 3$, $x\ne 0$
a,
$B=\Big( \dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}\Big) : \dfrac{3x^2}{x+3}$
$=\dfrac{-(x-3).(x+3)^2}{(x+3)(x-3)(x+3)} .\dfrac{x+3}{3x^2}$
$=-1.\dfrac{x+3}{3x^2}$
$=\dfrac{x+3}{-3x^2}$
b,
Ta có $3x^2>0$
$\Leftrightarrow -3x^2<0$
Để $B>0$ thì $x+3<0$
$\Leftrightarrow x<-3$