a) ĐK: $x\neq 0;x\neq \pm 1$
$P=\dfrac{1}{(x-1)^2}-\left [ \dfrac{x}{(x+1)(x-1)}-\dfrac{1}{x(x+1)(x-1)}\right ]:\dfrac{(x-1)^2}{x(x^2+1)}
\\=\dfrac{1}{(x-1)^2}-\dfrac{(x+1)(x-1)}{x(x+1)(x-1)}.\dfrac{x(x^2+1)}{(x-1)^2}
\\=\dfrac{1}{(x-1)^2}-\dfrac{x^2+1}{(x-1)^2}=\dfrac{1-x^2-1}{(x-1)^2}=\dfrac{-x^2}{(x-1)^2}$
$b)|x-1|=2\Rightarrow x=3\Rightarrow P=\dfrac{-3^2}{(3-1)^2}=\dfrac{-9}{4}$
$c)P=-1\Rightarrow \dfrac{-x^2}{(x-1)^2}=-1$
$\Rightarrow (x-1)^2=x^2\Leftrightarrow 1-2x=0\Leftrightarrow x=\dfrac{1}{2}$ (TM)