Số phần tử của tập hợp $A = \left\{ {{k^2} + 1|k \in \mathbb{N},\left| k \right| \le 2} \right\}$  là:

$8$

$10$

$12$

$7$

$x = y = 2.$    

$x = y = 2$ hoặc $x = 2,\,y = 5.$

$x = 2,\,y = 5.$

$x = 5,\,y = 2$ hoặc $x = y = 5.$

$A = \left\{ {x \in \mathbb{N}\left| {0 \le x \le 8} \right.}\right\}$, $B = \left\{ {3x \in \mathbb{N}\left| {1 \le x \le 5} \right.} \right\}$, $C = \left\{ {{2^x} \in \mathbb{N}\left| {1 \le x \le 4} \right.} \right\}$

$A = \left\{ {2x \in \mathbb{N}\left| {0 \le x \le 8} \right.} \right\},B = \left\{ {3x \in \mathbb{N}\left| {1 \le x \le 5} \right.} \right\},C = \left\{ {x \in \mathbb{N}\left| {1 \le x \le 16} \right.} \right\}$

$A = \left\{ {2x \in \mathbb{N}\left| {x \le 4} \right.} \right\},B = \left\{ {3x \in \mathbb{N}\left| {1 \le x \le 5} \right.} \right\},C = \left\{ {{2^x} \in \mathbb{N}\left| {x \le 4} \right.} \right\}$

$A = \left\{ {2x \in \mathbb{N}\left| {x \le 4} \right.} \right\},B = \left\{ {3x \in \mathbb{N}\left| {3 \le x \le 15} \right.} \right\},C = \left\{ {2x \in \mathbb{N}\left| {1 \le x \le 8} \right.} \right\}$

$2001$

$1001$

$1000$

$2000$