a,$\frac{(2x³+3x²)(2x+1)}{4x³-9x}$ =$\frac{x²(2x+3)(2x+1)}{x(4x²-9)}$ =$\frac{x²(2x+3)(2x+1)}{x(2x-3)(2x+3)}$ =$\frac{x(2x+1)}{2x-3}$
b,Để P=0⇒$\frac{x(2x+1)}{2x-3}$ =0
⇒x(2x+1)=0
⇒\(\left[ \begin{array}{l}x=0\\2x+1=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=-1/2\end{array} \right.\)