a. Xét $x\in A\cap (B\cup C)$

$\Rightarrow x\in A$ và $x\in B\cup C$

\(\Rightarrow \left\{\begin{matrix} x\in A\\ \left[\begin{matrix} x\in B\\ x\in C\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\in A\\ x\in B\end{matrix}\right.\\ \left\{\begin{matrix} x\in A\\ x\in C\end{matrix}\right.\end{matrix}\right.\Rightarrow x\in (A\cap B)\cup (A\cap C)(*)\)

Xét $x\in (A\cap B)\cup (A\cap C)$

$\Rightarrow x\in A\cap B$ hoặc $x\in A\cap C$

$\Rightarrow x\in A$ và $x\in B$ hoặc $x\in C$

Tức là: $x\in A\cap (B\cup C)(**)$

Từ $(*); (**)$ suy ra $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

b. Xét $x\in (A\setminus B)\setminus C$ bất kỳ

$\Rightarrow x\in A$ và $x\not\in B, x\not\in C$

Vì $x\in A, x\not\in C$ nên $x\in A\setminus C$

Do đó: $(A\setminus B)\setminus C\subset A\setminus C$

https://lazi.vn/edu/exercise/cho-cac-tap-hop-a-b-va-c-cmr-abc-abac

A vao day tham khao nhe !

~G#2k5~

\(A\backslash\left(B\cap C\right)=A\B\)\(\cup A\C\)

\(\left[{}\begin{matrix}x\in A\\\left\{{}\begin{matrix}x\in B\\x\in C\end{matrix}\right.\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\in A\\x\notin B\end{matrix}\right.\\\left\{{}\begin{matrix}x\in A\\x\notin C\end{matrix}\right.\end{matrix}\right.\)

F\(\cup\) G={b;c;e;g}\(\cup\) {c;d;e;f}

={b;c;d;e;f;g}

E\(\cap\) (F\(\cup\) G)={a;b;c;d} \(\cap\) {b;c;d;e;f;g}

={b;c;d}

E\(\cap\)F={a;b;c;d}\(\cap\) {b;c;e;g}

={b;c}

E\(\cap\) G={a;b;c;d}\(\cap\) {c;d;e;f}

={c;d}

(E\(\cap\)F)\(\cup\) (E\(\cap\) G)

={b;c}\(\cup\) {c;d}

={b;c;d}

Do đó: (E\(\cap\)F)\(\cup\) (E\(\cap\) G)=E\(\cap\) (F\(\cup\) G)