a) Let \(\sqrt{x}=a\Rightarrow x=a^2\)

\(\Rightarrow P=\left[\dfrac{a^2+3a+2}{\left(a+2\right)\left(a-1\right)}-\dfrac{a^2+a}{a^2-1}\right]:\left(\dfrac{1}{a+1}+\dfrac{1}{a-1}\right)\)

Compact \(P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

b) \(\dfrac{1}{P}-\dfrac{\sqrt{x}+1}{8}\ge1\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}\ge1\)

\(\Leftrightarrow\dfrac{16\sqrt{x}-\left(\sqrt{x}+1\right)^2}{8\left(\sqrt{x}+1\right)}\ge1\)

\(\Leftrightarrow16\sqrt{x}-\left(\sqrt{x}+1\right)^2\ge8\left(\sqrt{x}+1\right)\) (cause \(8\left(\sqrt{x}+1\right)>0\))

\(\Leftrightarrow\left(\sqrt{x}-3\right)^2\le0\Leftrightarrow x=9\)