Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
Cau\,1:\\
a)A = \left\{ {x \in N:\left| x \right| < 4} \right\} = \left\{ { - 3; - 2; - 1;0;1;2;3} \right\}\\
B = \left\{ {x \in Q:\left( {4{x^2} - x} \right)\left( {{x^2} + 3x - 4} \right) = 0} \right\}\\
\left( {4{x^2} - x} \right)\left( {{x^2} + 3x - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
4{x^2} - x = 0\\
{x^2} + 3x - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0 \in Q\\
x = \frac{1}{4} \in Q\\
x = 1 \in Q\\
x = - 4 \in Q
\end{array} \right. \Rightarrow B = \left\{ {0;\frac{1}{4};1; - 4} \right\}
\end{array}$
$\begin{array}{l}
Cau\,2:\\
a)A = \left[ {m;m + 2} \right],B = \left[ { - 1;2} \right]\\
A \subset B \Leftrightarrow - 1 \le m < m + 2 \le 2\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ge - 1\\
m + 2 \le 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ge - 1\\
m \le 0
\end{array} \right. \Leftrightarrow - 1 \le m \le 0
\end{array}$
