Đáp án:
$A.${$2$}
Giải thích các bước giải:
$A=${$x \in R | (2x-x^2)(2x^2-3x-2)=0$
$B=${$n \in N | 3 <n^2 <30$}
Ta có: $(2x-x^2)(2x^2-3x-2)=0 <=>$ \(\left[ \begin{array}{l}2x-x^2=0\\2x^2-3x-2=0\end{array} \right.\) $<=>$ \(\left[ \begin{array}{l}x=2\\x=0\\x=-\frac{1}{2} \end{array} \right.\)
$=>A=${$-\frac{1}{2};0;2$}
Ta có: $3<n^2 <=>$ \(\left[ \begin{array}{l}x < -\sqrt[]{3} \\x>\sqrt[]{3} \end{array} \right.\)
$n^2 <30 <=> -\sqrt[]{30} < x< \sqrt[]{30}$
$=>n \in (-\sqrt[]{30}; - \sqrt[]{3}) ∨ (\sqrt[]{3}; \sqrt[]{30})$
Mà $n \in N$
$=>n=${$-5;-4;-3;-2;2;3;4;5$}
$=>B=${$-5;-4;-3;-2;2;3;4;5$}
$=>A ∩ B =${$2$}
