1/ chứng minh rằng: a/ A giao (B\C) = ( A giao B)\ (A giao C) b/ A \ (B giao C) = (A \ B) hợp (A \ C)

Giải thích các bước giải:

a.Ta có:

$\eqalign{ & a)\,\,A \cap \left( {B\backslash C} \right) = \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \cr & \text{Lấy }x \in A \cap \left( {B\backslash C} \right) \cr & \Rightarrow \left\{ \matrix{ x \in A \hfill \cr x \in B\backslash C \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x \in A \hfill \cr x \in B \hfill \cr x \notin C \hfill \cr} \right. \Rightarrow \left\{ \matrix{ x \in A \cap B \hfill \cr x \notin A \cap C \hfill \cr} \right. \Rightarrow x \in \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \cr & \Rightarrow A \cap \left( {B\backslash C} \right) \subset \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \cr & \cr & \text{Lấy }x \in \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \cr & \Rightarrow \left\{ \matrix{ x \in A \cap B \hfill \cr x \notin A \cap C \hfill \cr} \right. \Rightarrow \left\{ \matrix{ x \in A \hfill \cr x \in B \hfill \cr \left[ \matrix{ \left\{ \matrix{ x \in A \hfill \cr x \notin C \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x \notin A \hfill \cr x \in C \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x \notin A \hfill \cr x \notin C \hfill \cr} \right. \hfill \cr} \right. \hfill \cr} \right. \Rightarrow \left\{ \matrix{ x \in A \hfill \cr x \in B \hfill \cr x \notin C \hfill \cr} \right. \Rightarrow \left\{ \matrix{ x \in A \hfill \cr x \in B\backslash C \hfill \cr} \right. \Rightarrow x \in A \cap \left( {B\backslash C} \right) \cr & \Rightarrow \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \subset A \cap \left( {B\backslash C} \right) \cr & \Rightarrow A \cap \left( {B\backslash C} \right) = \left( {A \cap B} \right)\backslash \left( {A \cap C} \right) \cr}$

b.Chứng minh: $A\setminus (B\cap C)\subseteq (A\setminus C)\cup (A\setminus B)$

Giả sử $x\in A\setminus (B\cap C)$

$\to \begin{cases} x\in A\\x\notin B\cap C\end{cases}$

$\to \begin{cases} x\in A\\x\notin B\quad hoặc\quad x\notin C\end{cases}$

$\to \begin{cases}x\in A\\ x\notin B\end{cases}$ hoặc $\begin{cases}x\in A\\ x\notin C\end{cases}$

$\to x\in (A\setminus B)$ hoặc $x\in (A\setminus C)$

$\to x\in (A\setminus B)\cup (A\setminus C)$

Chứng minh: $A\setminus (B\cap C)\supseteq (A\setminus C)\cup (A\setminus B)$

Giả sử $x\in (A\setminus C)\cup (A\setminus B)$

$\to x\in (A\setminus C)$ hoặc $x\in (A\setminus B)$

$\to \begin{cases} x\in A\\ x\notin C\end{cases}$ hoặc $\begin{cases} x\in A\\ x\notin B\end{cases}$ 

$\to \begin{cases} x\in A\\ x\notin C\quad hoặc\quad x\notin B\end{cases}$

$\to \begin{cases} x\in A\\ x\notin C\cap  B\end{cases}$

$\to x\in A\setminus (B\cap C)$