Đáp án:

$B=\left(\dfrac{x^2}{x^3-4x}-\dfrac{10}{5x+10}-\dfrac{1}{2-x}\right):\dfrac{6}{x^2-4x+4}\\ a)ĐKXĐ: \left\{\begin{array}{l} x \ne 0 \\ x\ne \pm 2\end{array} \right.\\ b)B=\dfrac{x-2}{x+2}\\ c)x \in \varnothing\\ d)\\ e)x \in \{-6;-4;-3;-1\}$

Giải thích các bước giải:

$B=\left(\dfrac{x^2}{x^3-4x}-\dfrac{10}{5x+10}-\dfrac{1}{2-x}\right):\dfrac{6}{x^2-4x+4}\\ a)ĐKXĐ: \left\{\begin{array}{l} x^3-4x \ne 0 \\ 5x+10 \ne 0 \\2-x \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x(x^2-4) \ne 0 \\ 5(x+2) \ne 0 \\2-x \ne 0 \\ \dfrac{6}{x^2-4x+4} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x(x-2)(x+2) \ne 0 \\ 5(x+2) \ne 0 \\2-x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ne 0 \\ x\ne \pm 2\end{array} \right.\\ b)B=\left(\dfrac{x^2}{x^3-4x}-\dfrac{10}{5x+10}-\dfrac{1}{2-x}\right):\dfrac{6}{x^2-4x+4}\\ =\left(\dfrac{x^2}{x(x-2)(x+2)}-\dfrac{10}{5(x+2)}+\dfrac{1}{x-2}\right):\dfrac{6}{(x-2)^2}\\ =\left(\dfrac{x}{(x-2)(x+2)}-\dfrac{2}{x+2}+\dfrac{1}{x-2}\right).\dfrac{(x-2)^2}{6}\\ =\left(\dfrac{x}{(x-2)(x+2)}-\dfrac{2(x-2)}{(x-2)(x+2)}+\dfrac{x+2}{(x-2)(x+2)}\right).\dfrac{(x-2)^2}{6}\\ =\dfrac{x-2(x-2)+x+2}{(x-2)(x+2)}.\dfrac{(x-2)^2}{6}\\ =\dfrac{6}{(x-2)(x+2)}.\dfrac{(x-2)^2}{6}\\ =\dfrac{x-2}{x+2}\\ c)B=-1\\ \Leftrightarrow \dfrac{x-2}{x+2}=-1\\ \Leftrightarrow \dfrac{x-2}{x+2}+1=0\\ \Leftrightarrow \dfrac{x-2+x+2}{x+2}=0\\ \Leftrightarrow \dfrac{2x}{x+2}=0\\ \Leftrightarrow x=0(L)\\ d)\\ e)B=\dfrac{x-2}{x+2} \in \mathbb{Z}\\ =\dfrac{x+2-4}{x+2} \in \mathbb{Z}\\ =1-\dfrac{4}{x+2} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{x+2} \in \mathbb{Z}\\ x\in \mathbb{Z} \Rightarrow  (x+2) \in Ư(4)\\ \Rightarrow x \in \{-6;-4;-3;-1;0;2\}\\ \text{Kết hợp điều kiện} \Rightarrow x \in \{-6;-4;-3;-1\}$