\[\begin{array}{l}
A = \frac{{x - 1}}{{4\sqrt x }};B = \frac{{1 + \sqrt x }}{{\sqrt x  - 1}} + \frac{{\sqrt x }}{{\sqrt x  + 1}} - \frac{{\sqrt x }}{{x - 1}},\left( {x > 0 \ne 1} \right)\\
B = \frac{{2x + 1}}{{x - 1}}\\
P = A.B = \frac{{x - 1}}{{4\sqrt x }}.\frac{{2x + 1}}{{x - 1}} = \frac{{2x + 1}}{{4\sqrt x }}
\end{array}\]

Ta có:\[\begin{array}{l}
P.4\sqrt x  \le \sqrt {4x - 4}  + x + 1\\
 \Leftrightarrow \frac{{2x + 1}}{{4\sqrt x }}.4\sqrt x  \le \sqrt {4x - 4}  + x + 1\\
 \Leftrightarrow 2x + 1 \le 2\sqrt {x - 1}  + x + 1\\
 \Leftrightarrow 2\sqrt {x - 1}  \ge x,\left( {x \ge 1} \right)\\
 \Leftrightarrow 4x - 4 \ge {x^2}\\
 \Leftrightarrow {x^2} - 4x + 4 \le 0\\
 \Leftrightarrow x = 2(t/m)
\end{array}\]

Giải thích:

\[\begin{array}{l}
{x^2} - 4x + 4 \le 0\\
 \Leftrightarrow {\left( {x - 2} \right)^2} \le 0
\end{array}\]

Theo ta biết thì \[{\left( {x - 2} \right)^2} \ge 0,\forall x\]

Như vậy ta có:\[\left\{ {\begin{array}{*{20}{c}}
{{{\left( {x - 2} \right)}^2} \ge 0}\\
{{{\left( {x - 2} \right)}^2} \le 0}
\end{array}} \right. \Leftrightarrow {\left( {x - 2} \right)^2} = 0 \Leftrightarrow x = 2\]