$\rm a) \\ ĐKXĐ:x\neq\pm3\\P= \dfrac{x+1}{3x-x^2} :( \dfrac{x+3}{3-x} - \dfrac{3-x}{3+x}- \dfrac{12x^2}{x^2-9})\\ P=\dfrac{-(x+1)}{x^2-3x}:(\dfrac{-(x+3)}{x-3}+\dfrac{-(3-x)}{x+3}-\dfrac{12x^2}{(x-3)(x+3)})\\P=\dfrac{-x-1}{x(x-3)}:(\dfrac{(-x-3)(x+3)+(x-3)(x-3)-12x^2}{(x-3)(x+3)})\\P=\dfrac{-x-1}{x(x-3)}: (\dfrac{-12x^2-12x}{(x-3)(x+3)})\\P=\dfrac{-x-1}{x(x-3)}.\dfrac{(x-3)(x+3)}{12x(-x-1)}\\P=\dfrac{x+3}{12x^2}\\\\b)\\|2x-1|=5 \Leftrightarrow \left[ \begin{array}{l}2x-1=5\\2x-1=-5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x=3(\text{không thoả mãn})\\x=-2(\text{thoả mãn})\end{array} \right.\\Với\ x=-2\ thì \ P=\dfrac{-2+3}{12.(-2)^2}=\dfrac{1}{48}\\Vậy \ P=\dfrac{1}{48} \ khi \ x=-2,\ P \ không \ xác \ định \ khi \ x=3\\\rm $
