Cho các tập hợp sau:

\(A = \left\{ {\left. {x \in \mathbb{R}} \right|\left( {{x^2} + 7x + 6} \right)\left( {{x^2} - 4} \right) = 0} \right\};\,B = \left\{ {\left. {x \in \mathbb{N}} \right|2x \le 8} \right\};\,C = \left\{ {\left. {2x + 1} \right|x \in \mathbb{Z}, - 2 \le x \le 4} \right\}\).

a) Tập hợp \(A\) có 3 phần tử.

b) \(A \cup B = \left\{ { - 6; - 2; - 1;0;1;2;3;4} \right\}\).

c) \(A \cap B = \left\{ 2 \right\}\).

d) \(A \cup C = \left\{ { - 6; - 3; - 2;2;3;5;7;9} \right\}\).

a) Sai. Ta có \(\left( {{x^2} + 7x + 6} \right)\left( {{x^2} - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}{x^2} + 7x + 6 = 0\\{x^2} - 4 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = - 6\\x = - 2\\x = 2\end{array} \right.\).

Vậy \(A = \left\{ { - 6; - 2; - 1;2} \right\}\).

b) Đúng. Ta có \(\left\{ {\begin{array}{*{20}{l}}{x \in \mathbb{N}}\\{2x \le 8}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{x \in \mathbb{N}}\\{x \le 4}\end{array} \Leftrightarrow x \in \left\{ {0;1;2;3;4} \right\}} \right.} \right.\). Vậy \(B = \left\{ {0;1;2;3;4} \right\}\).

Khi đó, \(A \cup B = \left\{ { - 6; - 2; - 1;0;1;2;3;4} \right\}\).

c) Đúng. Ta có \(A \cap B = \left\{ 2 \right\}\).

d) Sai. Ta có \(\left\{ {\begin{array}{*{20}{l}}{x \in \mathbb{Z}}\\{ - 2 \le x \le 4}\end{array} \Leftrightarrow x \in \left\{ { - 2; - 1;0;1;2;3;4} \right\}} \right.\). Suy ra \(C = \left\{ { - 3; - 1;1;3;5;7;9} \right\}\).

Khi đó, \(A \cup C = \left\{ { - 6; - 3; - 2; - 1;1;2;3;5;7;9} \right\}\).