Đáp án+Giải thích các bước giải:

$2)\\ a)A=\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{x^2-9} (x \ne \pm 3)\\ =\dfrac{5}{x+3}-\dfrac{2}{3-x}-\dfrac{3x^2-2x-9}{(x-3)(x+3)}\\ =\dfrac{5(x-3)}{(x+3)(x-3)}+\dfrac{2(x+3)}{(x-3)(x+3)}-\dfrac{3x^2-2x-9}{(x-3)(x+3)}\\ =\dfrac{5(x-3)+2(x+3)-(3x^2-2x-9)}{(x-3)(x+3)}\\ =\dfrac{9x-3x^2}{(x-3)(x+3)}\\ =\dfrac{-3x(x-3)}{(x-3)(x+3)}\\ =\dfrac{-3x}{x+3}\\ b)|x-2|=1\\ \Leftrightarrow \left[\begin{array}{l} x-2=1\\ x-2=-1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=3\\ x=1\end{array} \right.\\ x=1,A=\dfrac{-3.1}{1+3}=-\dfrac{3}{4}$

$x=3$ làm $A$ không xác định

$c)A \in \mathbb{Z}\\ \Leftrightarrow \dfrac{-3x}{x+3} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{-3x-9+9}{x+3} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{-3(x+3)+9}{x+3} \in \mathbb{Z}\\ \Leftrightarrow 3+\dfrac{9}{x+3} \in \mathbb{Z}\\ \Rightarrow \dfrac{9}{x+3} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (x+3) \in Ư(9)\\ \Leftrightarrow (x+3) \in \{\pm 1; \pm 3; \pm 9\}\\ \Leftrightarrow \left[\begin{array}{l} x+3=-9\\x+3=-3 \\x+3=-1\\ x+3=1 \\ x+3=3\\ x+3=9\end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=-12\\x=-6 \\x=-4\\ x=-2 \\ x=0\\ x=6\end{array} \right.\\ 3)\\ A=\left(\dfrac{x^3+1}{x^2-1}-\dfrac{x^2-1}{x-1}\right):\left(x+\dfrac{x}{x-1}\right)\\ a)\text{ĐKXĐ: } \left\{\begin{array}{l} x^2-1 \ne 0 \\ x-1 \ne 0  \\ \left(x+\dfrac{x}{x-1}\right) \ne 0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x^2 \ne 1 \\ x \ne 1  \\ \dfrac{x(x-1)+x}{x-1}\ne 0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x \ne \pm 1 \\ \dfrac{x^2}{x-1}\ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne \pm 1 \\ x \ne 0\end{array} \right.\\ A=\left(\dfrac{x^3+1}{x^2-1}-\dfrac{x^2-1}{x-1}\right):\left(x+\dfrac{x}{x-1}\right)\\ =\left(\dfrac{x^3+1}{(x-1)(x+1)}-\dfrac{(x^2-1)(x+1)}{(x-1)(x+1)}\right).\dfrac{x-1}{x^2}\\ =\dfrac{x^3+1-(x^2-1)(x+1)}{(x-1)(x+1)}.\dfrac{x-1}{x^2}\\ =\dfrac{-x^2+x+2}{(x-1)(x+1)}.\dfrac{x-1}{x^2}\\ =\dfrac{-x^2+x+2}{(x+1)x^2}\\ =\dfrac{-x^2-x+2x+2}{(x+1)x^2}\\ =\dfrac{-x(x+1)+2(x+1)}{(x+1)x^2}\\ =\dfrac{(2-x)(x+1)}{(x+1)x^2}\\ =\dfrac{2-x}{x^2}\\ b)A=3\\ \Leftrightarrow \dfrac{2-x}{x^2}=3\\ \Leftrightarrow \dfrac{2-x}{x^2}-3=0\\ \Leftrightarrow \dfrac{2-x-3x^2}{x^2}=0\\ \Leftrightarrow \dfrac{-3x^2-x+2}{x^2}=0\\ \Leftrightarrow \dfrac{-3x^2-3x+2x+2}{x^2}=0\\ \Leftrightarrow \dfrac{-3x(x+1)+2(x+1)}{x^2}=0\\ \Leftrightarrow \dfrac{(2-3x)(x+1)}{x^2}=0\\ \Leftrightarrow (2-3x)(x+1)=0\\ \Leftrightarrow \left[\begin{array}{l} x=-1\\ x=\dfrac{2}{3}\end{array} \right.\\ c)\circledast 2-x=0 \Leftrightarrow x=2 \Rightarrow A=0 (TM)\\ \circledast x \ne 2, A \in \mathbb{Z}\\ \Rightarrow |2-x| \ge x^2\\ \Rightarrow \left[\begin{array}{cc} 2-x \ge x^2&,x\le2  \\ x-2 \ge x^2 &, x > 2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x^2+x-2\le0&,x\le2  \\ x^2-x+2\le0 &, x > 2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} (x-1)(x+2)\le0&,x\le2  \\ x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\le0 &, x>2\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} (x-1)(x+2)\le 0 &,x\le2  \\ \left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\le 0 &, x> 2\end{array} \right.\\ $$ \Leftrightarrow (x-1)(x+2)\le0(x\le2)\\ \Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l} x-1 \ge0\\x+2 \le0 \end{array} \right. \\\left\{\begin{array}{l} x-1\le0\\x+2 \ge 0 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l} x \ge1\\x \le-2 \end{array} \right. \\\left\{\begin{array}{l} x\le1\\x \ge -2 \end{array} \right. \end{array} \right.\\ \Leftrightarrow -2 \le x \le1$

Kết hợp điều kiện và $x \in \mathbb{Z}, \Rightarrow x=-2 $

Thử lại thấy $x=-2$ thoả mãn

Vậy với $x=\pm 2$ thì $A \in \mathbb{Z}$

$4)\\ a)x^2-3x=0\\ \Leftrightarrow x(x-3)=0\\ \Leftrightarrow x=0; x=3\\ A(0)=\dfrac{-5}{-4}=\dfrac{5}{4}\\ A(3)=\dfrac{3-5}{3-4}=\dfrac{-2}{-1}=2\\ b)B=\dfrac{x+5}{2x}-\dfrac{x-6}{5-x}-\dfrac{2x^2-2x-50}{2x^2-10x}\\ =\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x(x-5)}\\ =\dfrac{(x+5)(x-5)}{2x(x-5)}+\dfrac{2x(x-6)}{2x(x-5)}-\dfrac{2x^2-2x-50}{2x(x-5)}\\ =\dfrac{(x+5)(x-5)+2x(x-6-(2x^2-2x-50)}{2x(x-5)}\\ =\dfrac{x^2-10x+25}{2x(x-5)}\\ =\dfrac{(x-5)^2}{2x(x-5)}\\ =\dfrac{x-5}{2x}\\ c)P=A:B=\dfrac{x-5}{x-4}:\dfrac{x-5}{2x}\\ =\dfrac{x-5}{x-4}.\dfrac{2x}{x-5}\\ =\dfrac{2x}{x-4}\\ =\dfrac{2x-8+8}{x-4}\\ =\dfrac{2(x-4)+8}{x-4}\\ =2+\dfrac{8}{x-4}\\ P \in \mathbb{Z} \Rightarrow \dfrac{8}{x-4} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (x-4) \in Ư(8)\\ \Leftrightarrow  (x-4) \in \{\pm 1; \pm 2; \pm 4; \pm 8\}\\ \Rightarrow x \in \{-4;0;2;3;5;6;8;12\}\\ \text{Kết hợp điều kiện} \Rightarrow x \in \{-4;2;3;6;8;12\}$