Đáp án:
$a)A=\{-1;0;3\}\\ b)B=\{1-\sqrt{6};-1;x=1+\sqrt{6}\}.$
Giải thích các bước giải:
$a)A=\{k^2-1|k \in \mathbb{Z},|k|<3\}\\ |k|<3 \Leftrightarrow -3 < k <3\\ k \in \mathbb{Z} \Rightarrow k \in \{-2;-1;0;1;2\}\\ k=-2, k^2-1=3\\ k=-1, k^2-1=0\\ k=0, k^2-1=-1\\ k=1, k^2-1=0\\ k=2, k^2-1=3\\ \Rightarrow A=\{-1;0;3\}\\ b)B=\{x \in \mathbb{R}|(x+1)(x^2-2x-5)=0\}\\ (x+1)(x^2-2x-5)=0\\ \Leftrightarrow \left[\begin{array}{l} x+1=0\\x^2-2x-5=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=-1\\x=1-\sqrt{6}\\x=1+\sqrt{6}\end{array} \right.\\ x \in \mathbb{R} \Rightarrow B=\{1-\sqrt{6};-1;x=1+\sqrt{6}\}.$
