Đáp án:
$\begin{array}{l}
B7)a)\\
Dkxd:x \ge 0;x \ne 1\\
P = \left( {\dfrac{1}{{\sqrt x - 1}} + \dfrac{{x - \sqrt x + 1}}{{x + \sqrt x - 2}}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} - \dfrac{{x - \sqrt x - 4}}{{x + \sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x + 2 + x - \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - x + \sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{x - 1 - x + \sqrt x + 4}}\\
= \dfrac{{x + 3}}{{\sqrt x + 3}}\\
b)P = 2\sqrt x - 1\\
\Rightarrow \dfrac{{x + 3}}{{\sqrt x + 3}} = 2\sqrt x - 1\\
\Rightarrow x + 3 = \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)\\
\Rightarrow x + 3 = 2x + 5\sqrt x - 3\\
\Rightarrow x + 5\sqrt x - 6 = 0\\
\Rightarrow \left( {\sqrt x - 1} \right)\left( {\sqrt x + 6} \right) = 0\\
\Rightarrow \sqrt x = 1\\
\Rightarrow x = 1\left( {ktm} \right)\\
Vay\,x \in \emptyset \\
c)P = \dfrac{{x + 3}}{{\sqrt x + 3}}\\
\Rightarrow P.\sqrt x + 3P = x + 3\\
\Rightarrow x - P.\sqrt x + 3 - 3P = 0\\
\Rightarrow \left\{ \begin{array}{l}
P > 0\\
3 - 3P > 0\\
\Delta \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 < P < 1\\
{P^2} - 4.\left( {3 - 3P} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
O < P < 1\\
{P^2} - 12 + 12P \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 < P < 1\\
{P^2} + 12P + 36 \ge 48
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 < P < 1\\
{\left( {P + 6} \right)^2} \ge 48
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 < P < 1\\
P + 6 \ge 4\sqrt 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
0 < P < 1\\
P \ge 4\sqrt 3 - 6
\end{array} \right.\\
\Rightarrow 4\sqrt 3 - 6 \le P < 1\\
\Rightarrow GTNN:P = 4\sqrt 3 - 6
\end{array}$
