Câu hỏi:

A. \(\left( {{\rm{A}}\backslash {\rm{B}}} \right) \cup \left( {{\rm{A}} \cap {\rm{C}}} \right) = \left\{ {{\rm{a}};{\rm{m}};{\rm{n}}} \right\}\);

B. \(\left( {{\rm{A}}\backslash {\rm{B}}} \right) \cup \left( {{\rm{A}} \cap {\rm{C}}} \right) = \left\{ {{\rm{a}};{\rm{c}};{\rm{m}};{\rm{n}}} \right\}\);

C. \(\left( {{\rm{A}}\backslash {\rm{B}}} \right) \cup \left( {{\rm{A}} \cap {\rm{C}}} \right) = \left\{ {{\rm{a}};{\rm{b}};{\rm{m}};{\rm{n}}} \right\}\);

D. \(\left( {{\rm{A}}\backslash {\rm{B}}} \right) \cup \left( {{\rm{A}} \cap {\rm{C}}} \right) = \left\{ {{\rm{a}};{\rm{n}}} \right\}\)

A. 0;

B. 1;

C. 0 và 1;

D. Không có

A. {2; 4};

B. {2};

C. {4; 5};

D. {3}

A. \[\left[ \begin{array}{l}{\rm{a}} \ge 3\\{\rm{a}} < 4\end{array} \right.\];

B. \[\left[ \begin{array}{l}{\rm{a}} > 3\\{\rm{a}} < - 4\end{array} \right.\];

C. \[\left[ \begin{array}{l}{\rm{a}} \ge 3\\{\rm{a}} \le - 4\end{array} \right.\];

D. \(\left[ \begin{array}{l}{\rm{a}} > 3\\{\rm{a}} \le - 4\end{array} \right.\)

A. \[\left[ \begin{array}{l}{\rm{a}} < \frac{5}{2}\\{\rm{a}} \ge - \frac{1}{3}\end{array} \right.\];

B. \(\left[ \begin{array}{l}{\rm{a}} \ge \frac{5}{2}\\{\rm{a}} < - \frac{1}{3}\end{array} \right.\);

C. \( - \frac{1}{3} \le {\rm{a}} < \frac{5}{2}\);

D. \[ - \frac{1}{3} \le {\rm{a}} \le \frac{5}{2}\]

A. 5;

B. 10;

C. 30;

D. 25

A. 25;

B. 10;

C. 45;

– π² < – 2 \( \Leftrightarrow \) π² < 4;

π < 4 \( \Leftrightarrow \) π² < 16;

\(\sqrt {23} < 5\,\, \Rightarrow \,\,2\sqrt {23} < 2.5\);

\(\sqrt {23} < 5\,\, \Rightarrow \,\, - 2\sqrt {23} > - 2.5\)

A. \(\exists x \in \mathbb{R},\,{x^2} + x + 5 \le 0\);

B. \[\forall x \in \mathbb{R},{x^2} + x + 5 \le 0\];

C. \[\exists \,x \in \mathbb{R},{x^2} + x + 5 < 0\];

D. \[\forall x \in \mathbb{R},\,{x^2} + x + 5 < 0\]