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TV ấn tượng nhất 2017
7 Tháng tám 2017
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1.
$ P = "\forall x \in \mathbb{R}: x^2 - 6x + 10 > 0" \\ x^2 - 6x + 10 = x^2 - 6x + 9 + 1 = (x - 3)^2 + 1 \geq 1 \Rightarrow (x - 3)^2 + 1 > 0 \\ \Rightarrow P\; đúng \\ \overline{P} = "\exists x \in \mathbb{R}: x^2 - 6x + 10 \leq 0"$
$ Q = "\exists x \in \mathbb{R}: \sqrt{3}x^2 + (1 + \sqrt{3})x + 1 = 0 " \\ \sqrt{3}x^2 + (1 + \sqrt{3})x + 1 = 0 \\ \Delta = (1 + \sqrt{3})^2 - 4 . \sqrt{3} . 1 = 1 + 2\sqrt{3} + 3 - 4\sqrt{3} = 4 - 2 \sqrt{3} > 0 \\ Phương\; trình\; có\; 2\; nghiệm\; phân\; biệt \\ \Rightarrow Q\; đúng \\ \overline{Q} = "\forall x \in \mathbb{R}: \sqrt{3}x^2 + (1 + \sqrt{3})x + 1 \neq 0" $
2.
a) $ \left \{ 1;2 \right \}; \left \{ 1;3 \right \}; \left \{ 1;4 \right \}; \left \{ 1;5 \right \}; \left \{ 2;3 \right \}; \left \{ 2;4 \right \}; \left \{ 2;5 \right \}; \left \{ 3;4 \right \}; \left \{ 3;5 \right \}; \left \{ 4;5 \right \} $
b) $ \left \{ 1;2 \right \}; \left \{ 1;2;3 \right \}; \left \{ 1;2;4 \right \}; \left \{ 1;2;5 \right \}; \left \{ 1;2;3;4 \right \}; \left \{ 1;2;3;5 \right \}; \left \{ 1;2;4;5 \right \}; \left \{ 1;2;3;4;5 \right \} $
3.
$ \\ A = \left \{ x \in \mathbb{R}: |2x - 1| \geq 1 \right \} \\ |2x - 1| \geq 1 \\ \Leftrightarrow 2x - 1 \geq 1\; hoặc\; 2x - 1 \leq -1 \\ \Leftrightarrow 2x \geq 2\; hoặc\; 2x \leq 0 \\ \Leftrightarrow x \geq 1\; hoặc\; x \leq 0 \\ \Rightarrow A = (-\infty ; 0] \cup [1; +\infty ) \\ B = \left \{ x \in \mathbb{R}: |2x + 1| \leq 5 \right \} \\ |2x + 1| \leq 5 \\ \Leftrightarrow -5 \leq 2x + 1 \leq 5 \\ \Leftrightarrow -6 \leq 2x \leq 4 \\ \Leftrightarrow -3 \leq x \leq 2\\ \Rightarrow B = [-3;2]\\ C = \left \{ x \in \mathbb{R}: a - 1 \leq x \leq a + 3 \right \} \\ \Rightarrow C = [a - 1;a + 3] \\ A \cup B = \mathbb{R}\\ A \cap B = [-3;0] \cup [1;2]\\ A \setminus B = (-\infty ;-3) \cup (2;+\infty )\\C_{\mathbb{R}}(A \cap B) = C_{\mathbb{R}}[-3;0] \cup [1;2] = (-\infty ;-3) \cup (0;1) \cup (2;+\infty ) $
$ C \subset B\\ \Rightarrow \left\{\begin{matrix}
-3 \leq a - 1\\
a + 3 \leq 2
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
-2 \leq a\\
a \leq -1
\end{matrix}\right. \\ \Leftrightarrow -2 \leq a \leq -1 $
4.
$ A \cap B \neq \varnothing \\ \Rightarrow -1 > 2a + 3\\ \Leftrightarrow -4 > 2a \\ \Leftrightarrow -2 > a $
$ A \setminus B \\ \Rightarrow A \subset B \\ \Rightarrow 3 < 2a + 3 \\ \Leftrightarrow 0 < 2a \\ \Leftrightarrow 0 < a $
 
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