Đáp án:
Giải thích các bước giải:
Câu 1:
ĐKXĐ: $-x^2+5x+6\ge 0\Leftrightarrow (x+1)(6-x)\ge 0\Leftrightarrow -1\le x\le 6$
Vậy D=[-1;6]
Câu 2:
$\left\{ \begin{gathered} {x^2} - 5x + 6 > 0 \hfill \\ 3 - 2x \leqslant 0 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} \left[ \begin{gathered} x < 2 \hfill \\ x > 3 \hfill \\ \end{gathered} \right. \hfill \\ x \geqslant \frac{3}{2} \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} \frac{3}{2} \leqslant x < 2 \hfill \\ x > 3 \hfill \\ \end{gathered} \right.$
Chọn C
Câu 3: Đề bài tương đương với $(m-3)x^2+2(2m-3)x+5m-6>0$ có nghiệm đúng $\forall x\in\mathbb{R}$
$\Leftrightarrow \left\{ \begin{gathered} m = 3 \hfill \\ 2(2m - 3) = 0 \hfill \\ 5m - 6 > 0 \hfill \\ \end{gathered} \right.(L)$ hoặc
$\left\{ \begin{gathered} m - 2 > 0 \hfill \\ \Delta ' < 0 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} m > 2 \hfill \\ (2m - 3){}^2 - (m - 2)(5m - 6) < 0 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} m > 2 \hfill \\ - {m^2} + 4m - 3 < 0 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} m > 2 \hfill \\ \left[ \begin{gathered} m < 1 \hfill \\ m > 3 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \Leftrightarrow m > 3$. Chọn D
Câu 4: ĐKXĐ:
$\left\{ \begin{gathered} {x^2} + 3x \geqslant 0 \hfill \\ x + 2 > 0 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} x > - 2 \hfill \\ \left[ \begin{gathered} x \ge 0 \hfill \\ x \le - 3 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right. \Leftrightarrow x > 0$
Vậy
$D = {\text{[}}0; + \infty )$
