Đáp án A đúng vì nếu lấy \(x \in A\backslash B \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x \in A}\\{x \notin B}\end{array}} \right. \Rightarrow x \in A\)
\( \Rightarrow A\backslash B \subset A\)
Đáp án B đúng vì nếu lấy \(x \in B\backslash A \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x \in B}\\{x \notin A}\end{array}} \right. \Rightarrow x \in B\)
\( \Rightarrow B\backslash A \subset B\)
Đáp án C đúng vì nếu lấy \(x \in A\backslash B \cup B\backslash A \Rightarrow \left[ {\begin{array}{*{20}{c}}{x \in A\backslash B}\\{x \in B\backslash A}\end{array}} \right. \)
\(\Rightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \in A}\\{x \notin B}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x \in B}\\{x \notin A}\end{array}} \right.}\end{array}} \right. \Rightarrow \left[ {\begin{array}{*{20}{c}}{x \in A}\\{x \in B}\end{array}} \right. \)
\(\Rightarrow x \in A \cup B\)
Đáp án D sai. Phản ví dụ:
Giả sử \(A = \left\{ 1 \right\};B = \left\{ {1,2} \right\}\).
Ta có:\(1 \in A \cup B\) nhưng
\(\left\{ {\begin{array}{*{20}{c}}{1 \notin A\backslash B}\\{1 \notin B\backslash A}\end{array}} \right. \Rightarrow 1 \notin A\backslash B \cup B\backslash A\)
Đáp án đúng: D
