$\begin{array}{l}A = \left(\dfrac{1}{x-2} - \dfrac{2x}{4 - x^2} + \dfrac{1}{2+ x}\right)\cdot \left(\dfrac{2}{x} - 1\right)\\\text{ĐKXĐ: }x \ne 0; \, x \ne \pm 2\\a) \, A = \left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right)\cdot \left(\dfrac{2}{x} - 1\right)\\=\left[\dfrac{x+2}{(x-2)(x+2)}+\dfrac{2x}{(x-2)(x+2)}+\dfrac{x-2}{(x-2)(x+2)}\right]\cdot \left(\dfrac{2-x}{x}\right)\\=\left[\dfrac{x+2 + 2x + x -2}{(x-2)(x+2)}\right]\cdot\dfrac{(x-2)}{-x}\\=\dfrac{4x}{(x+2)(-x)}\\=\dfrac{-4}{x+2}\\b)\, \text{Ta có:}\\2x^2+x = 0\\\Leftrightarrow x(2x + 1) = 0\\\Leftrightarrow \left[ \begin{array}{l}x=0\\x=-\dfrac{1}{2}\end{array} \right.\\Với \,\, x = 0: \text{không thỏa ĐKXĐ} \Rightarrow \text{A không xác định}\\Với \, \, x = -\dfrac{1}{2} \Rightarrow A = \dfrac{-4}{-\dfrac{1}{2} + 2} = -\dfrac{8}{3}\\c) \, A = \dfrac{1}{2}\\\Leftrightarrow \dfrac{-4}{x +2} = \dfrac{1}{2}\\ \Leftrightarrow x + 2 = -8 \\\Leftrightarrow x = -10\\\text{Vậy x = -10}\\d) \, A = \dfrac{-4}{x+2}\\A \in \Bbb Z^{+} \Leftrightarrow \dfrac{-4}{x+2} \in \Bbb Z^+\\\Leftrightarrow \begin{cases} (x+2) \in Ư(-4)\\x+2 <0\end{cases}\\\Leftrightarrow x+2 = \left\{-4;-2;-1\right\}\\\text{Bảng giá trị:}\\\begin{array}{|l|r|}\hline x+2 & -4 & -2&-1 \\ \hline x & -6& -4 &-3\\\hline \end{array} \\\text{Vậy x={-6;-4;-3} để A nguyên dương}\end{array}$
