Câu hỏi
Cho các tập hợp: \(A = \left\{ {\left. {x \in \mathbb{R}} \right|\,\left( {{x^2} + 7x + 6} \right)\left( {{x^2} - 4} \right) = 0} \right\}\); \(B = \left\{ {x \in \mathbb{N}|\,2x \le 8} \right\}\); \(C = {\rm{\{ }}2x + 1|\,x \in \mathbb{Z};\,\,\, - 2 \le x \le 4{\rm{\} }}{\rm{.}}\) Tìm \(\left( {A \cup C} \right)\backslash B.\)
- A \((A \cup C)\backslash B = \left\{ { - 3; - 1;5;7;9} \right\}\)
- B \((A \cup C)\backslash B = \left\{ { - 6; - 3; - 2; - 1;5} \right\}\)
- C \((A \cup C)\backslash B = \left\{ { - 6; - 3; - 2; - 1;5;7;9} \right\}\)
- D \((A \cup C)\backslash B = \left\{ { - 6; - 3;5;7;9} \right\}\)
Phương pháp giải:
Giải các phương trình, tìm \(A,B,C\) từ đó tìm \(A \cup C\) để tìm \(\left( {A \cup C} \right)\backslash B.\)
Lời giải chi tiết:
\( \bullet \) Ta có: \(\left( {{x^2} + 7x + 6} \right)\left( {{x^2} - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}{x^2} + 7x + 6 = 0\\{x^2} - 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = - 6\\x = - 2\\x = 2\end{array} \right..\)
Vậy \(A = \left\{ { - 6; - 2; - 1;2} \right\}.\)
\( \bullet \) Ta có: \(\left\{ \begin{array}{l}x \in \mathbb{N}\\2x \le 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \in \mathbb{N}\\x \le 4\end{array} \right. \Leftrightarrow x \in \left\{ {0;\,\,1;\,\,2;\,\,3;\,\,4} \right\}.\)
Vậy \(B = \left\{ {0;\,\,1;\,\,2;\,\,3;\,\,4} \right\}.\)
\( \bullet \) Ta có: \(\left\{ \begin{array}{l}x \in \mathbb{Z}\\ - 2 \le x \le 4\end{array} \right.\, \Leftrightarrow x \in \left\{ { - 2; - 1;\,\,0;\,\,1;\,\,2;\,\,3 ;\,\,4} \right\} \Rightarrow \left( {2x + 1} \right) \in \left\{ { - 3; - 1;\,1;\,3;\,\,5;\,\,7;\,\,9} \right\}.\)
\( \Rightarrow C = \left\{ { - 3; - 1;\,1;\,3;\,5;\,7;\,9} \right\}\)
Ta có: \(A \cup C = \left\{ { - 6; - 3; - 2; - 1;1;2;3;5;7;9} \right\}\)
\( \Rightarrow \left( {A \cup C} \right)\backslash B = \left\{ { - 6; - 3; - 2; - 1;\,5;\,7;\,9} \right\}.\)
Chọn C.
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