Trong tam giác $ABC$, ta có.

$A \cap B = B$

$A \cup B = \;A$  

${C_A}B = \{ 0;4\} $

$B\backslash A = \{ 0;4\} $ .

\(\left( {1;1} \right) \in S\).

\(\left( {\dfrac{{\sqrt 2 }}{2};0} \right) \in S\).

\(\left( {1; - 2} \right) \notin S\).

\(\left( {1;0} \right) \notin S\).

0

1

2

Vô số

$A = \left\{ {-1;1} \right\}$

$A = \left\{ {-\sqrt 2 ;-1;1;\sqrt 2 } \right\}$

$A = \left\{ {-1} \right\}$

$A = \left\{ 1 \right\}$

$\left( {-\infty ;3} \right) \cup \left[ {3; + \infty } \right) = R$

$R\backslash \left( {-\infty ;0} \right) = {R_ + }$  

$R\backslash (0; + \infty ) = {R_ - }$

$R\backslash \left( {0; + \infty } \right) = R_ - ^*$

\( - \dfrac{1}{2}x - y + 1 \le 0\)

\( - \dfrac{1}{2}x - y + 1 > 0\)

\( - \dfrac{1}{2}x - y + 1 \ge 0\)

\( - \dfrac{1}{2}x - y + 1 < 0\)

$\emptyset  \subset A$.

$1 \in A$

$\{ 1;2\}  \subset A$

$2 = A$.