Đáp án:
$\begin{array}{l}
a)Dkxd:\dfrac{{ - 3}}{{x + 4}} \ge 0\\
\Rightarrow x + 4 < 0\\
\Rightarrow x < - 4\\
Vậy\,x < - 4\\
b)\\
1)x = 2\\
\Rightarrow {2^2} - 2.\left( {m + 1} \right).2 + 2m + 10 = 0\\
\Rightarrow 4 - 4m - 4 + 2m + 10 = 0\\
\Rightarrow 2m = 10\\
\Rightarrow m = 5\\
Khi:m = 5\\
\Rightarrow {x^2} - 12x + 20 = 0\\
\Rightarrow {x^2} - 2x - 10x + 20 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x - 10} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 10
\end{array} \right.\\
Vậy\,khi:m = 5;x = 2;x = 10\\
2)\Delta ' > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 2m - 10 > 0\\
\Rightarrow {m^2} + 2m + 1 - 2m - 10 > 0\\
\Rightarrow {m^2} > 9\\
\Rightarrow \left[ \begin{array}{l}
m > 3\\
m < - 3
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 2m + 10
\end{array} \right.\\
P = 10{x_1}{x_2} + x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} + 8{x_1}{x_2}\\
= 4{\left( {m + 1} \right)^2} + 8.\left( {2m + 10} \right)\\
= 4.\left( {{m^2} + 2m + 1 + 4m + 20} \right)\\
= 4\left( {{m^2} + 6m + 21} \right)\\
= 4.\left( {{m^2} + 6m + 9 + 12} \right)\\
= 4.{\left( {m + 3} \right)^2} + 48 \ge 48\\
\Rightarrow P \ge 48\\
\Rightarrow GTNN:P = 48\\
Khi:m = - 3
\end{array}$
