Solve b*1/x+2<1/(x-2)^2

$$b \cdot \frac { 1 } { x + 2 } < \frac { 1 } { ( x - 2 ) ^ { 2 } }$$

Solve for x

$$\left\{\begin{matrix}x\in \mathrm{R}\setminus 2,-2\text{, }&b=0\\x\in (-2,2)\cup (2,\infty)\text{, }&b<-\frac{1}{16}\text{ or }(b\geq -\frac{1}{16}\text{ and }b<0)\\x\in \mathrm{R}\setminus (2,-2\cup (-\infty,-\frac{-4b+\sqrt{16b+1}-1}{2b}]\text{, }b\leq 0\cup [\frac{4b+\sqrt{16b+1}+1}{2b},\infty)\cup [-2,-\frac{-4b+\sqrt{16b+1}-1}{2b}]\text{, }b>0)\text{, }&b>0\\x\in (\frac{4b+\sqrt{16b+1}+1}{2b},-\frac{-4b+\sqrt{16b+1}-1}{2b})\text{, }&b>-\frac{1}{16}\text{ and }b<0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b>\frac{x+2}{\left(x-2\right)^{2}}\text{, }&x<-2\\b<\frac{x+2}{\left(x-2\right)^{2}}\text{, }&x>-2\text{ and }|x|\neq 2\end{matrix}\right.$$