Đáp án:
19.A
20.D
Giải thích các bước giải:
$\begin{array}{l}
19)y' = 3{x^2} + 2mx - \left( {4m + 9} \right)(*)\\
ham\,so\,dong\,bien\,\forall x \in R \Leftrightarrow y' \ge 0\forall x \in R\\
\Leftrightarrow 3{x^2} + 2mx - \left( {4m + 9} \right) \ge 0\,\,\forall x \in R\\
\Leftrightarrow \Delta ' \le 0\\
\Leftrightarrow {m^2} + 12m + 27 \le 0\\
\Leftrightarrow \left( {m + 3} \right)\left( {m + 9} \right) \le 0\\
\Leftrightarrow - 9 \le m \le - 3 \Rightarrow m \in \left\{ { - 9; - 8;...; - 3} \right\}\,co\,7\,gtri\,cua\,m\\
20.\\
Xet\,pt\,hoanh\,do\,giao\,diem\,\\
\Leftrightarrow - 5x + 21 = \frac{{x + 3}}{{x - 2}}\\
\Rightarrow \left( { - 5x + 21} \right)\left( {x - 2} \right) = x + 3\\
\Rightarrow - 5{x^2} + 10x + 21x - 42 = x + 3\\
\Rightarrow - 5{x^2} + 30x - 45 = 0\\
\Rightarrow {x^2} - 6x + 9 = 0\\
\Rightarrow {x_0} = 3 \Rightarrow {y_0} = - 5{x_0} + 21 = 6\\
\Rightarrow {x_0} + {y_0} = 9
\end{array}$
