Đáp án:

$\begin{array}{l}
1)a)Dkxd:x \ge 0;x\# 1;x\# \dfrac{1}{4}\\
B = \left( {\dfrac{{x\sqrt x  + x + \sqrt x }}{{x\sqrt x  - 1}} - \dfrac{{\sqrt x  + 3}}{{1 - \sqrt x }}} \right).\dfrac{{x - 1}}{{2x + \sqrt x  - 1}}\\
 = \left( {\dfrac{{\sqrt x \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 1}}} \right)\\
.\dfrac{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {2\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \left( {\dfrac{{\sqrt x }}{{\sqrt x  - 1}} + \dfrac{{\sqrt x  + 3}}{{\sqrt x  - 1}}} \right).\dfrac{{\sqrt x  - 1}}{{2\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 3}}{{\sqrt x  - 1}}.\dfrac{{\sqrt x  - 1}}{{2\sqrt x  - 1}}\\
 = \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}}\\
B < 0\\
 \Leftrightarrow \dfrac{{2\sqrt x  + 3}}{{2\sqrt x  - 1}} < 0\\
 \Leftrightarrow 2\sqrt x  - 1 < 0\\
 \Leftrightarrow \sqrt x  < \dfrac{1}{2}\\
 \Leftrightarrow x < \dfrac{1}{4}\\
Vậy\,0 \le x < \dfrac{1}{4}\\
b){x^2} - \left( {2m + 5} \right).x + 2m + 1 = 0\\
\Delta  > 0\\
 \Leftrightarrow {\left( {2m + 5} \right)^2} - 4\left( {2m + 1} \right) > 0\\
 \Leftrightarrow 4{m^2} + 20m + 25 - 8m - 4 > 0\\
 \Leftrightarrow 4{m^2} + 12m + 21 > 0\left( {ld} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 5\\
{x_1}{x_2} = 2m + 1
\end{array} \right.\\
Khi:{x_1},{x_2} > 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
2m + 5 > 0\\
2m + 1 > 0
\end{array} \right.\\
 \Leftrightarrow m > \dfrac{{ - 1}}{2}\\
P = \left| {\sqrt {{x_1}}  - \sqrt {{x_2}} } \right|\\
 = \sqrt {{{\left( {\sqrt {{x_1}}  - \sqrt {{x_2}} } \right)}^2}} \\
 = \sqrt {{x_1} - 2\sqrt {{x_1}{x_2}}  + {x_2}} \\
 = \sqrt {2m + 5 - 2\sqrt {2m + 1} } \\
 = \sqrt {2m + 1 - 2\sqrt {2m + 1}  + 1 + 4} \\
 = \sqrt {{{\left( {\sqrt {2m + 1}  - 1} \right)}^2} + 4}  \ge \sqrt 4  = 2\\
 \Leftrightarrow P \ge 2\\
 \Leftrightarrow GTNN:P = 2\\
Khi:\sqrt {2m + 1}  = 1\\
 \Leftrightarrow m = 0\left( {tmdk} \right)\\
Vậy\,m = 0
\end{array}$