Đáp án:
$ a)A=\dfrac{1}{2-x}\\ b) x =\pm \dfrac{1}{2}\\ x=\dfrac{1}{2};A=\dfrac{2}{3}\\ x=-\dfrac{1}{2};A=\dfrac{2}{5}\\ c) x=\dfrac{3}{2}\\ d) x>2\\ e) \left[\begin{array}{l} x=3\\x=1\end{array} \right.$
Giải thích các bước giải:
$A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\\ \text{ĐKXĐ}: \left\{\begin{array}{l}x^3-4x \ne 0\\ 6-3x \ne 0 \\ x+2 \ne 0\\ \left(x-2+\dfrac{10-x^2}{x+2}\right) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}x(x^2-4) \ne 0\\ 3(2-x) \ne 0 \\ x \ne -2\\\dfrac{(x-2)(x+2)+10-x^2}{x+2} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}x(x-2)(x+2) \ne 0\\ 2-x \ne 0 \\ x \ne -2\\\dfrac{6}{x+2} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}x \ne 0\\ x \ne \pm 2\end{array} \right.\\ a)A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\\ =\left(\dfrac{x^2}{x(x-2)(x+2)}-\dfrac{6}{3(x-2)}+\dfrac{1}{x+2}\right):\dfrac{6}{x+2}\\ =\left(\dfrac{3x^2}{3x(x-2)(x+2)}-\dfrac{6x(x+2)}{3x(x-2)(x+2)}+\dfrac{3x(x-2)}{3x(x-2)(x+2)}\right).\dfrac{x+2}{6}\\ =\dfrac{-18x}{3x(x-2)(x+2)}.\dfrac{x+2}{6}\\ =\dfrac{1}{2-x}\\ b)|x|=\dfrac{1}{2} \Rightarrow x =\pm \dfrac{1}{2}\\ x=\dfrac{1}{2};A=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}\\ x=-\dfrac{1}{2};A=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{1}{\dfrac{5}{2}}=\dfrac{2}{5}\\ c)A=2\Leftrightarrow \dfrac{1}{2-x}=2\\ \Leftrightarrow \dfrac{1}{2-x}-2=0\\ \Leftrightarrow \dfrac{1-2(2-x)}{2-x}=0\\ \Leftrightarrow \dfrac{2x-3}{2-x}=0\\ \Leftrightarrow 2x-3=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ d)A<0\\ \Leftrightarrow \dfrac{1}{2-x}<0\\ \Leftrightarrow 2-x<0\\ \Leftrightarrow x>2\\ e)\dfrac{1}{2-x} \in \mathbb{Z}; x \in \mathbb{Z}; \\ \Rightarrow (2-x) \in Ư(1)\\ \Leftrightarrow (2-x) \in \{\pm1\}\\ \Leftrightarrow \left[\begin{array}{l} 2-x=-1\\2-x=1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=3\\x=1\end{array} \right.$
