Đáp án:

$ a)D=\dfrac{4}{x-1}\\ b)D(2)=4\\ c) x<1, x \ne -2,x \ne -1 , x \ne 0\\ d) x=3\\ e) x \in \{-3;2;3;5\}\\ f)x \in \varnothing\\ g)max_{(2-D-D^2)}=\dfrac{9}{4} \Leftrightarrow  x=-7$

Giải thích các bước giải:

$D=\left(\dfrac{x}{x+2}+\dfrac{8x+8}{x^2+2x}-\dfrac{x+2}{x}\right):\left(\dfrac{x^2-x-3}{x^2+2x}+\dfrac{1}{x}\right)\\ ĐKXĐ:  \left\{\begin{array}{l} x+2 \ne 0 \\ x^2+2x \ne 0  \\ x \ne 0 \\ \left(\dfrac{x^2-x-3}{x^2+2x}+\dfrac{1}{x}\right) \ne 0 \end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ne -2 \\ x(x+2) \ne 0  \\ x \ne 0 \\ \left(\dfrac{x^2-x-3}{x(x+2)}+\dfrac{1}{x}\right) \ne 0  \end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ne -2 \\ x \ne 0 \\ \left(\dfrac{x^2-x-3}{x(x+2)}+\dfrac{(x+2)}{x(x+2)}\right) \ne 0 \end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ne -2 \\ x \ne 0 \\ \dfrac{x^2-x-3+x+2}{x(x+2)} \ne 0 \end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ne -2 \\ x \ne 0 \\ \dfrac{x^2-1}{x(x+2)} \ne 0 \end{array} \right.\\ \Leftrightarrow  \left\{\begin{array}{l} x \ne -2 \\ x \ne 0 \\ x \ne \pm 1\end{array} \right.\\ a)D=\left(\dfrac{x}{x+2}+\dfrac{8x+8}{x^2+2x}-\dfrac{x+2}{x}\right):\left(\dfrac{x^2-x-3}{x^2+2x}+\dfrac{1}{x}\right)\\ =\left(\dfrac{x}{x+2}+\dfrac{8x+8}{x(x+2)}-\dfrac{x+2}{x}\right):\left(\dfrac{x^2-x-3}{x(x+2)}+\dfrac{1}{x}\right)\\ =\left(\dfrac{x^2}{x+2}+\dfrac{8x+8}{x(x+2)}-\dfrac{(x+2)^2}{x(x+2)}\right):\left(\dfrac{x^2-x-3}{x(x+2)}+\dfrac{(x+2)}{x(x+2)}\right)\\ =\dfrac{x^2+8x+8-(x+2)^2}{x(x+2)}:\dfrac{x^2-x-3+x+2}{x(x+2)}\\ =\dfrac{4x+4}{x(x+2)}:\dfrac{x^2-1}{x(x+2)}\\ =\dfrac{4x+4}{x(x+2)}.\dfrac{x(x+2)}{x^2-1}\\ =\dfrac{4(x+1)}{x(x+2)}.\dfrac{x(x+2)}{(x-1)(x+1)}\\ =\dfrac{4}{x-1}\\ b)D=\dfrac{4}{x-1}\\ x(x-2)-x+2=0\\ \Leftrightarrow x(x-2)-(x-2)=0\\ \Leftrightarrow ( x-1)(x-2)=0\\ \Leftrightarrow  \left[\begin{array}{l} x=1\\ x=2\end{array} \right.$

$x=1$ làm $D$ không xác định nên không tính được $D(1)$

$D(2)=\dfrac{4}{2-1}=4\\ c)D<0\\ \Leftrightarrow \dfrac{4}{x-1}<0\\ \Leftrightarrow x-1<0\\ \Leftrightarrow x<1$

Kết hợp điều kiện 

$\Rightarrow x<1, x \ne -2,x \ne -1 , x \ne 0\\ d)D=2\\ \Leftrightarrow \dfrac{4}{x-1}=2\\ \Leftrightarrow \dfrac{4}{x-1}-2=0\\ \Leftrightarrow \dfrac{4-2(x-1)}{x-1}=0\\ \Leftrightarrow \dfrac{6-2x}{x-1}=0\\ \Leftrightarrow 6-2x=0\\ \Leftrightarrow x=3\\ e)D \in \mathbb{N^*}\\ \Leftrightarrow \dfrac{4}{x-1}  \in \mathbb{N^*}\\ \Rightarrow (x-1) \in Ư(4)\\ \Rightarrow x \in \{-3;-1;0;2;3;5\}$

Kết hợp điều kiện $\Rightarrow x \in \{-3;2;3;5\}$

$f)(x^2-1)D=4x-2\\ \Leftrightarrow (x-1)(x+1)\dfrac{4}{x-1}=4x-2\\ \Leftrightarrow 4(x+1)=4x-2\\ \Leftrightarrow 4x+4=4x-2\\ \Leftrightarrow 4=-2(\text{Vô lí})\\ g)2-D-D^2\\ =-D^2-D+2\\ =-D^2-D-\dfrac{1}{4}+\dfrac{9}{4}\\ =-\left(D+\dfrac{1}{2}\right)^2+\dfrac{9}{4} \le \dfrac{9}{4} \ \forall \ D$

Dấu "=' xảy ra $\Leftrightarrow D+\dfrac{1}{2}=0$

$\Leftrightarrow D=-\dfrac{1}{2}\\ \Leftrightarrow  \dfrac{4}{x-1}=-\dfrac{1}{2}\\ \Leftrightarrow  \dfrac{4}{x-1}+\dfrac{1}{2}=0\\ \Leftrightarrow  \dfrac{8+x-1}{2(x-1)}=0\\ \Leftrightarrow  \dfrac{x+7}{2(x-1)}=0\\ \Leftrightarrow  x+7=0\\ \Leftrightarrow  x=-7$