a) Ta có: \(x \in \left( {A \cap B} \right) \cup A \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x \in A \cap B}\\{x \in A}\end{array}} \right. \) \(\Leftrightarrow x \in A\)
Vậy \((A \cap B) \cup A = A\)
b) Ta có: \(x \in \left( {A \cup B} \right) \cap B \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \in A \cup B}\\{x \in B}\end{array}} \right.\) \( \Leftrightarrow x \in B\)
Vậy \((A \cup B) \cap B = B\)
c) Ta có: \(x \in \left( {A\backslash B} \right) \cup B \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x \in A\backslash B}\\{x \in B}\end{array}} \right. \) \(\Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x \in A}\\{x \notin B}\end{array}} \right.}\\{x \in B}\end{array}} \right. \Leftrightarrow x \in A \cup B\)
Vậy \((A\backslash B) \cup B = A \cup B\)
d) Ta có: \(x \in \left( {A\backslash B} \right) \cap \left( {B\backslash A} \right) \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{x \in A\backslash B}\\{x \in B\backslash A}\end{array}} \right.\) \( \Leftrightarrow x \in \emptyset \)
Vậy \((A\backslash B) \cap (B\backslash A) = \emptyset \)
