Toán cho ba tập hợp:A;B;C Chứng minh rằng :A giao (B \C)=(A giao B)\C 30/09/2021 By Eliza cho ba tập hợp:A;B;C Chứng minh rằng :A giao (B \C)=(A giao B)\C
Giải thích các bước giải: $\begin{array}{l} + )x \in A \cap \left( {B\backslash C} \right) \Leftrightarrow \left\{ \begin{array}{l}x \in A\\x \in B\backslash C\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \in A\\x \in B\\x\not \in C\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \in \left( {A \cap B} \right)\\x\not \in C\end{array} \right. \Leftrightarrow x \in \left( {A \cap B} \right)\backslash C\\ \Rightarrow A \cap \left( {B\backslash C} \right) \subset \left( {A \cap B} \right)\backslash C\left( 1 \right)\\ + )x \in \left( {A \cap B} \right)\backslash C \Leftrightarrow \left\{ \begin{array}{l}x \in A \cap B\\x\not \in C\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \in A\\x \in B\\x\not \in C\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \in A\\x \in B\backslash C\end{array} \right. \Leftrightarrow x \in A \cap \left( {B\backslash C} \right)\\ \Rightarrow \left( {A \cap B} \right)\backslash C \subset A \cap \left( {B\backslash C} \right)\left( 2 \right)\\\left( 1 \right),\left( 2 \right) \Rightarrow A \cap \left( {B\backslash C} \right) = \left( {A \cap B} \right)\backslash C\end{array}$ Trả lời
Giải thích các bước giải:
$\begin{array}{l}
+ )x \in A \cap \left( {B\backslash C} \right) \Leftrightarrow \left\{ \begin{array}{l}
x \in A\\
x \in B\backslash C
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \in A\\
x \in B\\
x\not \in C
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \in \left( {A \cap B} \right)\\
x\not \in C
\end{array} \right. \Leftrightarrow x \in \left( {A \cap B} \right)\backslash C\\
\Rightarrow A \cap \left( {B\backslash C} \right) \subset \left( {A \cap B} \right)\backslash C\left( 1 \right)\\
+ )x \in \left( {A \cap B} \right)\backslash C \Leftrightarrow \left\{ \begin{array}{l}
x \in A \cap B\\
x\not \in C
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \in A\\
x \in B\\
x\not \in C
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \in A\\
x \in B\backslash C
\end{array} \right. \Leftrightarrow x \in A \cap \left( {B\backslash C} \right)\\
\Rightarrow \left( {A \cap B} \right)\backslash C \subset A \cap \left( {B\backslash C} \right)\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow A \cap \left( {B\backslash C} \right) = \left( {A \cap B} \right)\backslash C
\end{array}$