a) Ta có $\forall x,\,\,x \in A \cup C \Leftrightarrow \left[ {\begin{array}{*{20}{c}} {x \in A}\\ {x \in C} \end{array}} \right.$
Với $x\in A$ vì $A\subset B\Rightarrow x\in B\Rightarrow x\in B\cup D$
Suy ra $A\cup C\subset B\cup D$.
b) Ta có $\forall x,\,\,x \in A \cap C \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {x \in A}\\ {x \in C} \end{array}} \right. \Rightarrow x \in A$
Vì $A\subset B\Rightarrow x\in B$
Suy ra $A\cap C\subset B$.
c) $\forall x,\,\,x \in {C_B}A \cup A \Leftrightarrow \left[ {\begin{array}{*{20}{c}} {x \in {C_B}A}\\ {x \in A} \end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}} {\left\{ {\begin{array}{*{20}{c}} {x \in B}\\ {x \notin A} \end{array}} \right.}\\ {x \in A} \end{array}} \right. \Leftrightarrow x \in B$
Suy ra ${{C}_{B}}A\cup A=B$
