Đáp án:
$\begin{array}{l}
B = \left( {\dfrac{{x + 3\sqrt x - 2}}{{x - 9}} - \dfrac{1}{{\sqrt x + 3}}} \right).\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{x + 3\sqrt x - 2 - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
2)b)\\
M\left( {1; - 1} \right) \in \left( d \right)\\
\Rightarrow - 1 = m - 1 + \dfrac{1}{2}{m^2} + m\\
\Rightarrow {m^2} + 4m = 0\\
\Rightarrow m\left( {m + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\\
m = - 4
\end{array} \right.\\
c)\dfrac{1}{2}{x^2} = \left( {m - 1} \right)x + \dfrac{1}{2}{m^2} + m\\
\Rightarrow {x^2} - 2\left( {m - 1} \right)x - {m^2} - 2m = 0\\
\Rightarrow \Delta ' = {\left( {m - 1} \right)^2} + {m^2} + 2m\\
= {m^2} - 2m + 1 + {m^2} + 2m\\
= 2{m^2} + 1 > 0
\end{array}$
=> chúng luôn cắt nhau tại 2 điểm phân biệt
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = - {m^2} - 2m
\end{array} \right.\\
Khi:x_1^2 + x_2^2 + 6{x_1}{x_2} > 2019\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} + 4{x_1}{x_2} > 2019\\
\Rightarrow 4{\left( {m - 1} \right)^2} + 4.\left( { - {m^2} - 2m} \right) > 2019\\
\Rightarrow 4{m^2} - 8m + 4 - 4{m^2} - 8m > 2019\\
\Rightarrow - 16m > 2015\\
\Rightarrow m < \dfrac{{ - 2015}}{{16}}
\end{array}$
