Đáp án:
\(\begin{array}{l}
a)A = \left\{ {\dfrac{1}{9}} \right\}\\
b)B = \left\{ \emptyset \right\};B \subset A\\
C = \left\{ {\dfrac{1}{9}} \right\};C \subset A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
\dfrac{{2021}}{{3\sqrt x + 2020}}\\
Do:x \ge 0 \to 3\sqrt x + 2020 > 0\\
\to \dfrac{{2021}}{{3\sqrt x + 2020}} > 0\\
Do:x \ge 0 \to 3\sqrt x \ge 0\\
\to 3\sqrt x + 2020 \ge 2020\\
\to \dfrac{{2021}}{{3\sqrt x + 2020}} \le \dfrac{{2021}}{{2020}}\\
\to \dfrac{{2021}}{{2020}} \ge \dfrac{{2021}}{{3\sqrt x + 2020}} > 0\\
Do:\dfrac{{2021}}{{3\sqrt x + 2020}} \in Z\\
\to \dfrac{{2021}}{{3\sqrt x + 2020}} = 1\\
\to 3\sqrt x + 2020 = 2021\\
\to 3\sqrt x = 1\\
\to \sqrt x = \dfrac{1}{3}\\
\to x = \dfrac{1}{9}\\
a)A = \left\{ {\dfrac{1}{9}} \right\}\\
b)B = \left\{ \emptyset \right\};B \subset A\\
C = \left\{ {\dfrac{1}{9}} \right\};C \subset A
\end{array}\)
