\(a,A=\dfrac{x+2}{\left(x-2\right)^2}\left(x\ne2\right)\\ \left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-1\left(n\right)\end{matrix}\right.\\ \Rightarrow A=\dfrac{2-1}{\left(-1-2\right)^2}=\dfrac{1}{9}\\ b,B=\dfrac{x+2}{x}+\dfrac{1}{x-2}+\dfrac{6-x^2}{x^2-2x}\left(x\ne0;x\ne\pm2\right)\\ B=\dfrac{x^2-4+x+6-x^2}{x\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ \to P=\dfrac{A}{B}=\dfrac{x+2}{\left(x-2\right)^2}\cdot\dfrac{x\left(x-2\right)}{x+2}=\dfrac{x}{x-2}\\ c,P< 1\Leftrightarrow\dfrac{x}{x-2}-1< 0\\ \Leftrightarrow\dfrac{2}{x-2}< 0\Leftrightarrow x-2< 0\Leftrightarrow x< 2\\ \Leftrightarrow x< 2;x\ne0;x\ne-2\)
