$\begin{array}{l}P = \left(\dfrac{x-1}{x-3} + \dfrac{x-3}{x+3} - \dfrac{4x - 2}{9 -x^2} \right):\left(1 - \dfrac{x + 1}{3 + x}\right)\\ a)\,\,ĐKXĐ: \, 9 - x^2 \ne 0\Leftrightarrow x \ne \pm 3\\ P = \left[\dfrac{(x-1)(x + 3)}{x-3} + \dfrac{(x-3)^2}{x+3} + \dfrac{4x - 2}{(x-3)(x+3)} \right]:\left(\dfrac{x + 3}{x+3} - \dfrac{x + 1}{3 + x}\right)\\ \to P = \dfrac{x^2 + 3x - x - 3 + x^2 - 6x + 9 + 4x - 2}{(x-3)(x+3)} : \dfrac{2}{x+3}\\ \to P = \dfrac{2x^2 + 4}{(x-3)(x+3)}\cdot\dfrac{x+3}{2}\\ \to P = \dfrac{x^2 + 2}{x - 3}\\ b)\,Ta\,\,có:\\ x^2 - 3|x| = 0\\ \Leftrightarrow |x|^2 - 3|x| = 0\\ \Leftrightarrow \left[\begin{array}{l}|x| = 0\\|x| = 3\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = 0\\x = \pm 3 \quad (loại)\end{array}\right.\\ Với\,\,x= 0\,\,ta\,\,được:\\ P = \dfrac{0^2 + 2}{0 - 3} = - \dfrac{2}{3}\\ c)\,\,P = \dfrac{x^2 + 2}{x- 3} = \dfrac{x^2 - 6x + 9 +6x - 18 + 11}{x - 3}\\ \to P = \dfrac{(x-3)^2 + 6(x-3) + 11}{x-3}\\ \to P = x - 3 + 6 + \dfrac{11}{x-3}\\ \to P = x + 3 + \dfrac{11}{x-3}\\ P \in \Bbb Z \Leftrightarrow \dfrac{11}{x-3} \in \Bbb Z\\ \Leftrightarrow \begin{cases}x - 3 \in Ư(11)\\x \in \Bbb Z\end{cases}\\ \text{Ta có bảng giá trị:}\\\begin{array}{|l|r|} \hline x -3& -11&-1&1&11\\ \hline \quad x&-8&2&4&14\\ \hline \end{array}\\ Vậy\,\,x = \left\{-8;2;4;14\right\}\end{array}$
