Đáp án:
$\begin{array}{l}
1a)Dkxd:x \ge 0;x\# 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 + 2} \right)\left( {\sqrt x - 1} \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}}\\
P = 2\sqrt x - 1\\
\Leftrightarrow \dfrac{{x + 3}}{{\sqrt x + 3}} = 2\sqrt x - 1\\
\Leftrightarrow x + 3 = 2x + 6\sqrt x - \sqrt x - 3\\
\Leftrightarrow x + 5\sqrt x - 6 = 0\\
\Leftrightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\left( {tmdk} \right)\\
x = 9\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 4;x = 9\\
b)P = \dfrac{{x + 3}}{{\sqrt x + 3}} = \dfrac{{x + 3\sqrt x - 3\sqrt x - 9 + 12}}{{\sqrt x + 3}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 12}}{{\sqrt x + 3}}\\
= \sqrt x - 3 + \dfrac{{12}}{{\sqrt x + 3}}\\
= \sqrt x + 3 + \dfrac{{12}}{{\sqrt x + 3}} - 6\\
Do:\sqrt x + 3 > 0\\
Theo\,Co - si:\\
\sqrt x + 3 + \dfrac{{12}}{{\sqrt x + 3}} \ge 2\sqrt {\left( {\sqrt x + 3} \right).\dfrac{{12}}{{\sqrt x + 3}}} \\
\Leftrightarrow \sqrt x + 3 + \dfrac{{12}}{{\sqrt x + 3}} \ge 2\sqrt {12} = 2.2\sqrt 3 = 4\sqrt 3 \\
\Leftrightarrow \sqrt x + 3 + \dfrac{{12}}{{\sqrt x + 3}} - 6 \ge 4\sqrt 3 - 6\\
\Leftrightarrow GTNN:P = 4\sqrt 3 - 6\\
Khi:\sqrt x + 3 = \dfrac{{12}}{{\sqrt x + 3}}\\
\Leftrightarrow \sqrt x + 3 = 2\sqrt 3 \\
\Leftrightarrow \sqrt x = 2\sqrt 3 - 3\\
\Leftrightarrow x = {\left( {2\sqrt 3 - 3} \right)^2} = 21 - 12\sqrt 3 \left( {tmdk} \right)\\
Vậy\,GTNN:P = 4\sqrt 3 - 6\\
2)a)A\left( { - 2; - 1} \right) \in \left( P \right)\\
\Leftrightarrow - 1 = a.{\left( { - 2} \right)^2}\\
\Leftrightarrow a = - \dfrac{1}{4}\\
\Leftrightarrow \left( P \right):y = - \dfrac{1}{4}{x^2}\\
b){x_B} = 4\\
\Leftrightarrow {y_B} = - \dfrac{1}{4}x_B^2 = - \dfrac{1}{4}{.4^2} = - 4\\
\Leftrightarrow B\left( {4; - 4} \right)\\
A\left( { - 2; - 1} \right)\\
Goi:AB:y = a.x + b\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4 = a.4 + b\\
- 1 = a.\left( { - 2} \right) + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - \dfrac{1}{2}\\
b = - 2
\end{array} \right.\\
Vậy\,AB:y = - \dfrac{1}{2}x - 2\\
c)Goi:\left( d \right):y = a.x + b\\
Do:\left( d \right)//AB\\
\Leftrightarrow \left( d \right):y = - \dfrac{1}{2}.x + b\left( {b\# - 2} \right)\\
Xet:\dfrac{{ - 1}}{4}{x^2} = - \dfrac{1}{2}x + b\\
\Leftrightarrow {x^2} - 2x + 4b = 0\\
\Delta ' = 0\\
\Leftrightarrow 1 - 4b = 0\\
\Leftrightarrow b = \dfrac{1}{4}\left( {tmdk} \right)\\
Vậy\,\left( d \right):y = - \dfrac{1}{2}x + \dfrac{1}{4}
\end{array}$