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}$